6
Work, Energy and Power
Introductory Exercise 6.1 1.
→ → → Work done by F = F⋅ x
2.
→ → → Work done by F = F ⋅ l
→ N
→
→
→
T
W
T
N
F
→
→
→
T
β
→ l
→ → → |F| = F , |N| = N → → |W| = W , |T| = T → |l| = l
→
– x
→ F
→ → → |F| = F , |N|= N , |W|= W → → |T|= T a T and |x|= x → → cos π = |F ||x| co
= − F x → → → Work done by| by |N|= N ⋅ x = |N||x| cos π =0 → → = W⋅x
→ → = |W|| l | cos π + α
→ Work done by W
2
= |W|⋅|x| cos π →
2
= T x
→
→ → → Work do done by by |W|= W ⋅ l
2
=0 → → → Work done by T = T ⋅ x → → = |T||x|cos 0
→
= |F|⋅| l | cos π = − F l → → → Work done by N = N ⋅ l → → = |N|⋅| l | cos π = 0 2
→ →
→
α
→ W
= − W l sin α → → → Work Work don done e by |T|= T ⋅ l →
→
= |T|| l |cos β = T l cos β
Work, Energy and Power | 3.
→ W
→ → − T = m g 4
Substituting value of N from Eq. (ii) in Eq. (i).
µ (W − F sin 45° ) = F cos 45° 1 1 1 or W − F = F
g a = 4
→ T
4
or
2
F
W −
2
2
=4
F 2
5 F = W 2 2 2 or F = W = mg 5 5 → → Work done by force F force F = F ⋅ s → → = |F||s|cos 45° = F s 1 2 = 2 mg s 1 = mgs 5 2 5 1.8 × 10 × 2 or
→
W = mg
→ l
=
→ → |g |= g and | l |= l
µ N = F cos 45° F sin sin 45° N
→ → Work done by friction = µ N ⋅ s → → = µ |N||s|cos π
= − µ Ns = − F cos cos 45° s 7.2 J = − 7.2 J → → Work done by gravity = g ⋅ s → → = |g|⋅|s| cos π 2
…(i) F
5
= 7.2 J 7.2 J
→ → → m g |T|= m g − ∴ 4 3 → = m g 4 → → Work done by string = T ⋅ l → → = |T|| l |cos π → → = 3 m g l cos π 4 = − 3 mgl 4 4.
5.
=0 F = mg sin sin 45°
F
2ms –1
45°
F
F cos cos 45°
µN W
mg sin 45°
s
mg cos cos 45° 45°
→ → N = |N|, W = |W| → → → s = |s |, F = |F|, g = |g |
N
109
+ F sin 45° = W
= 1 × 10 × …(ii)
=5
1 2
2N
Displacement of lift in 1s
=2m
110
| Mechanics-1 → → Work Work done done by force force of fricti friction on ( F ) = F ⋅ s → → = |F||s|cos 45° = F s cos 45° = 5 2 ⋅ 2 ⋅ 1 = 10 Nm 10 Nm 2
6.
Total work-done by spring on both masses m
Work done = Area under the curve
7.
F (N) (N) 10
– 4
–2
A2
A3
A4 2
A1
x (m)
4
m – 10
x 0 m
m
= A1 + A2 + A3 + A4 − 2 × − 10 − 2 × 10 = + + [10 × 2] 2 2 2 × 10 + 2 30 Nm = 30 Nm
= PE of the spring when stretched by 2 x0 = 1 k (2x0 )2 2
= 2 k x20 ∴ Work done 2 k x20 = = k x20
by spring on each mass
2
Introductory Exercise 6.2 1.
a=
∴
−20 ms −1 2 s F
= − 10 ms −2
B
= ma = 2 kg × − 10 ms −2 = − 20 N s = Area under curve curve 1 = × 25 × 20 ms −1
work done by F + + Work done by f p = Work done by F (pseudo force)
= ( mas) + ( − mas) = 0 As P As P is is at rest , ∆ K = 0 ∴ ∆ K = W (Work -Energy theorem)
According to A to A (inertial (inertial frame) P
m a
Q
Acceleration of P = a
∴
Note
In iner inertial frames one has to also to con sider work-done due to pseudo forces, while ap plying plying Work-energy Work-energy theo theorem.
3.
Force on P = ma
Work done over s over s displacement displacement = mas
∴
= u2 + 2as ( u = 0) = 2as 1 1 mas ∴ Gain in KE = mv2 = m 2 as = ma Now,
v2
2
∴
a
According to B to B (non-inertial (non-inertial frame)
= 20 m 20 m ∴ Work done = F s = ( −20 N) (20 m) = − 400 Nm A
F
Q
2
2.
f P P
2 α F = ma = m
2
∆ K = W (Work -Energy theorem)
v=α x dv a= = α 1 dx dt 2 x dt 2 1 α =α α x = 2 x 2 2
∴
W =
mα b 2 2
Work, Energy and Power | 4.
by F ∆ K = Work done by F + Work done by gravity
or
7.
4m
x2 2 m x = v0 A
− 1 mv20 = − 2
⇒
F = = 80 N
g F
111
A
(a) If T = mg, the block will not get accel acceler erated ated to gain KE. The value of T must be greater that Mg that Mg..
5 g
5.
= 80 ⋅ 4 ⋅ cos 0 + 5 g ⋅ 4 ⋅ cos π = 320 + ( − 200) or K f − K i = 120 J or K f = 120 J (as K (as K i = 0) Change in KE = Work done R (1 (1 – cosθ)
mg R sin sin
θ
T
F
T
T
Hand
θ
Mg
mg
Ans. False ∴ Ans. False
R mg
O
(b) As some some negative negative work work will be done by Mg, Mg, the work done by T will be more that 40 J. Ans. False ∴ Ans. False
1 mv2 = mgR (1 − cos θ) + mgR sin θ 2 ⇒ v = 2 gR (1 − cos θ + sin θ) 6.
∆ K = W x 1 or 0 − mv20 = ∫ − 0 2
(c) Pulling Pulling force force F will will always be equal to T , as T is is there only because of pulling. Ans. True ∴ Ans. True (d) Work done by gravity will be negative Ans. False Ans. False
Ax dx
Introductory Exercise 6.3 1.
Spring is having its natural length.
In Fig. 1
A
T
T
T
T A
x
m A
B
m
T T B m
Ground
Now, for the block B to B to just leave contact with ground
Ground
Fig. 2
= mg 2m A g = mg m m A = kx
i.e.,
mg Fig. 1
In Fig. 2 A is A is released. A goes A goes down by x . x . Spring get extended by x by x.. Decrease in PE of A of A is is stored in spring as its PE. 1 ∴ mAg x = k x2 2
⇒
2
112 2.
| Mechanics-1
1 mv2 2
∴
=
v=
i.e., i.e., 3.
l 2 l mg 2
Decrease in PE = mg
gl
4.
OA = 50 cm C 20 20 cm B cm B
1 1 mv2C = × 500 × 0.0924 2 2 500 or vc = × 0.0924 = 2.15m s −1 100 m Work done by man = gh + Mgh 2 = m + M gh 2 or
40 cm A 5.
30 cm
When block of man M man M goes goes down by x by x,, the spring gets extended by x by x.. Decrease in PE of man M man M is is stored in spring as its PE.
O
collar is at A at A)) ∴ Extension in spring (when collar = 50 cm − 10 cm = 0.4 m
m
m
Extension in spring (collar is at B at B)) 37°
= 30 cm − 10 cm = 20 cm = 0.2 m
T
T
KE of collar at B at B
T
= PE of spring − PE of spring (collar at a t A)
T T
(collar at a t B)
mg sin 37°
= 1 × K × [(0.4)2 − (0.2)2 ] or
1 2 mv B = m
or
v B =
(30)2
37° mg
(Collar at C)
= × 500 × [(0.4) − (0.26) 2
Mgx =
or
kx
2
1 2 kx 2
= 2Mg
k x = mg sin 37 ° + µ mg cos 37 ° 3 3 4 2 Mg = mg + mg 5 4 5 3 M = m 5
or
= PE of spring − PE of spring 1 2
∴
For the block of man m to just slide
+ (20)2 − 10] cm = 0.26 m
KE of collar at C (Collar at A at A))
T
cos 37° µmg cos
2 1 × 500 × 12 2 500 × 0.12 = 2.45 s −1 10
Extension in spring (collar arrives at C)
=[
x
or ]
Introductory Exercise 6.4 1.
Velocity at time t = 2 s v = g t = 10 × 2 = 20 ms −1
Power = Force × velocity
2.
Velocity at time = a t =
∴vav =
= mgv = 1 × 10 × 20 = 200 W Pav
=
F ⋅t m
Ft (acceleration being constant) 2m F 2 t F × vav = 2m
Work, Energy and Power | As at t = 0, KE = 0, c = 0
Instantaneous power
= Force × instantaneous velocity 2 = F ⋅ Ft = F t
3.
m m Energy Power = Time KE P = ⇒ KE = P ⋅ t t 1 ∴ mv2 = Pt 2 2 Pt or v= m ds or = 2 P ⋅ t1 / 2 dt m 2P or ds = t1 / 2 dt m
∫
2 P t 3 / 2 m 3 / 2
P = 2t
∫
KE = t2 1 2 i.e., i.e., mv2 = t2 or v = t 2 m 2.0 + 2 t Pav = =t 2
dx2
=2
∴ Equilibrium is stable. 6. F = x − 4 For equilibrium, F equilibrium, F = = 0 i.e., i.e., x − 4 = 0 i.e., i.e., x = 4 m dU As, F As, F = −
+c
dx
∫
=t +c
− 20 + ( x − 2)2 at x = 2. PE( = U is U ) is minimum at x at x = 2 m ∴ Equilibrium position is at x dU = 2 ( x − 2) d2U
KE = P dt = 2t dt 2
U=
dx
At t = 0, s = 0, ∴ c = 0 8 P 3 / 2 Thus, s = t 9m 4.
∴
5.
∫
s =
or
113
dU = − ( x − 4) dx d2U =−1 dx2
∴
Thus, equilibrium is unstable.
AIEEE Corner Subjective Questions (Level I) (a) Work done by a constant force 1.
3.
(a) Work done by a constant force Work done by applied force = F s cos0
= 40 × 2 = 80 Nm Work done by force of gravity = mgscos mgs cos π
= 2 × 10 × 2 × − 1 = − 40 Nm 2.
→ r21
→
→
…(i) …(ii)
Solving Eq. (i) and Eq. (ii), m − m2 a= 1 g m1 + m2
−1 g 4+1 m1 = 4 kg and m2 = 1 kg =
= r2 − r1
4
^
^
= (2 i + 3 j − 4 k) − (1 i + 4 j + 6 k) ^
− T = m1a T − m2 g = m2 a m1 g
^
^
^
= 3 g
= ^i − ^j − 10 k^
s = ut + ^
^
= (6 i − 2 j + k) ⋅ ( i − j − 10 k) = 6 + 2 − 10 = − 2 Nm ^
^
^
T
T T
5
→ → Work done = F ⋅ r21 ^
a
1 2 at 2
a T m2g m1g
114
| Mechanics-1
= 1 3 g 22 2 5 12 m = 12 m
= 2 × 10 × 2 sin 60° = 20 3 Nm = 34.6 Nm 34.6 Nm
Work done by gravity on 4 kg block
Work done by force of friction
= 4 g × 12 cos 0 480 Nm = 480 Nm
= f s cos π = µ Ns cos π = − µ ( mg cos θ) s = − 1 (2 × 10 × cos 60° ) × 2
Solving Eq. (i) and Eq. (ii), 2 m1m2 T = g m1 + m2 = 2 × 4 g 5
2
10 Nm = − 10 Nm (b) Work done by a variable force.
= 16 N
W
6.
Work done by string on 1 kg block
= 16 × 12 cos 0 = 192 Nm 4.
x
F = = 16 N
F
45°
W
7.
= ∫ F dx x
=2
= ∫ x x = 4
4
dx x2 x = 2 = x = 4 4 x −2 dx x = 2 x − 2 + 1 = 4 − 2 + 1 x = 4 x = 2 1 =− 4 x x = 4
s = 2.2 m
∫
mg
= 16 × 2 ⋅ 2 ×
1 2
= 24.9 Nm Work done by normal force
= N s cos 90° = 0
= − 4 1 − 1 = − 1 Nm 2 4
Work done by force of gravity
= mgs cos 90° = 0
(c) Work done by by area under under F- x graph
Total work done on the block
5.
=− 4
x2 = − 2 2 x = 2 = − [( −4)2 − (2)2 ] = − 12 Nm 12 Nm
Work done by applied force = F s cos 45° N
= ∫ F dx x = − 4 = ∫ x = 2 − 2 x dx
= 24.9 + 0 + 0 = 24.9 Nm 24.9 Nm Work done by gravity = mgs sin θ N
8.
(a) W = 3
15 Nm × (5 − 10) = − 15 Nm
F x (N) 3
µ N
f =
x (m) (m)
s mg
θ = 60 60 °
2
4
(b) W = 3
6
8
10
12
× (1 0 − 5) = + 15 Nm
14
16
115
Work, Energy and Power | (c) W =
3
× (12 − 10)
∴ minimum PE = ( 4 − 4)2 − 16 = − 16J Now, (KE) max + (PE) min =
= 3 Nm 2 (10 − 4) + (12 − 0) (d) W = ×3 9.
2 = + 27 Nm (3 − 2) + (3
(a) W =
− 0)
2
or
×2
+ ( − 16) = − 4 (KE) max = 12 J = KE max = 12 J 12 J (KE) max
or (c) PE max
F x (N)
PEmax 2
O
(d) W = 4 Nm + 0 Nm + ( − 1) Nm
∴
2
3
4
5
6
7
2
= 3 Nm. Conservative force field and Potential Energy. d 10. F = − U dr = − d Ar−1 dr = ( −) ( − A) r−1−1= A2 r U = ( x − 4)2
− 16 (at x = 6.0m) ∴ PE (at x = (6 − 4)2 − 16 = − 12 J KE (at KE (at x x = 6.0 m) = 8 J (a) ∴ Total mechanical energy = ( − 12) + (8) = − 4 J (b) KE will be maximum maximum where, where, PE, is minimum. For U to to be minimum, dU =0 dx d i.e., i.e., [( x − 4)2 − 16] = 0 dx or 2 ( x − 4) = 0 or
x = 4 m
x
x = = 4 KEmax
= 4 Nm (b) W = 0 Nm 1 (c) W = (6 − 4) ( − 1 − 0) = − 1 Nm
1
– 1
11.
PEmax
− 16 or 12 = ( x − 4) − 16 or x2 − 8 x − 12 = 0 x = 4 + 2 3 ⇒ = 4−2 3 and 2 (d) U = ( x − 4) − 16 dU = 2 ( x − 4)
x (m) (m) O
Total
mechanical energy
U = ( x − 4)2
2
or
F x
=
dx − dU = dx F x = 8
− 2 ( x − 4) − 2x
(e) F (e) F x = 0 i.e., i.e., or
− 2 x = 0 x = 4 m
8
Kinetic energy and Work-energy theorem p2 12. K = 2m 2 p + p 2 ∴ K ′ = 2m ( K ′ is the KE when momentum p is increased by 50%) 50%) 9 p2 or K ′ = 4 2m 9 or K ′ = K 4 9 5 or K′ − K = K − K = K 4 4 K ′ − K 5 ∴ = K 4 = 5 × 100% = 125 % 4
116 13.
| Mechanics-1
p = (2 mK )1 / 2
∴
16.
p′ = [2m ( K
+ 1% of K ) ]
1 / 2
( p′ is the momentum when KE i.e., K is increased by 1%) 1%)
i.e.,
p′ = p +
14.
= 4 ms −2
F
= m d 2s
dt
∴
17.
F
N
2.5
T
m
= m( g − a) = 30 (10 − 0.04) = 298.8 N Work done by chain = T s cos π Nm = − (298.8 × 2) Nm = − 597.6 Nm
= 2.5 − x2
∴ F decreases decreases as x as x increases increases and F and F is is zero when x when x = 2.5 m Thus, work will be +ive from x = 0 to x = 2.5 m an so KE will be maximum at x = 2.5 m. KE (at x (at x = 2.5 m) 2.5 = ∫ 0 (2.5 − x2 ) dx
2
T
23 3
(b) Position of maximum KE
− 2.2 + 10 = 14 m ∴ ∆ s = 14 m − 10 m = 4 m Work = F ∆s =8N× 4m = 32 Nm
i.e.,
2
x3 2 0
2.33 Nm = 2.33 Nm (1 Nm = 1 J) J) ∴ ∆ K = 2.33 J i.e., KE i.e., KE (at x (at x = 2) − KE (at x = 0) = 2.33 J 2.33 J ∴ KE (at x = 2) = 2.33 J
dt
= (0.4) = 0.16 = 0.04 m s −2 2 s 2×2 4 mg − T =a
mg
= ∫ F dx
= (2.5 × 2) −
s (at t = 2 s) = 2.2
v2
(a) W
= 2.5 x −
2
a=
a
2
s (at t = 0 s) = 10 m
15.
− F = ma F = m ( g − a) = 5 (10 − 2.5) = 37.5 N
F (Push (Push of air)
= ∫ 0 (2.5 − x2 ) dx
2
= 2 × 4= 8
2 × 20
= F s cos π = − (37.5 × 20) = − 750 Nm
∴ Increase in momentum = 0.5%. 0.5%. s = (2t2 − 2t + 10) m ds = 4t − 2 2
2.5 ms−2 = 2.5
Work done by push of air
1 % of p 2
dt d2 s
102
Now, mg
1
= [2 mK (1 + 1%) ]2 1 / 2 = (2 mK )1 / 2 1 + 1 100 1 / 2 p′ = p 1 + 1 100 = p 1 + 1 × 1 2 100 = p 1 + 1 % 2
a =
a
x3 = 2.5 x − 3 0 2.5 2.5 = 2.5 2.5 − 3 = 2 × 2.5 3
2.635 J = 2.635 J
2.5
Work, Energy and Power | 18.
F
a=
− mg
Adding Eq. (i) and Eq. (ii),
m g 10
= F − g m 11 F= mg 10
or or
∴ a=
m1
m1 g
F
= u2 + 2as v2 = 2as v = 2as = 2×2×1 = 2 ms −1
i.e., or
mg
(a) ∴ Work done on astronaut by F by F = 11mg × 15 10 = 11 × 72 × 10 × 15 10
20.
T
10
T
astronaut astronaut
by
T
a
= (11642.4) + ( − 10584) = 1058.4 Nm 1058.4 Nm KE = 1058.4 J ∴ KE 1 (d) mv2 = 1058.4 2
72
= 5.42
and
T
T
or
− 50 = 5a T ′ = 2T T ′ T = 2
…(i) …(ii)
a m2
50 (N)
T'
300 (N)
i.e.,
= m2 a m1 g − T = m1 a T
Also, i.e.,
T a
1m
…(ii)
T ′ − 100 = 10a 300 a 300 − T ′ = 10
…(iii)
300 − T ′ = 30a
…(iv)
Solving Eq. (iii) and Eq. (iv), a = 5 ms −2
m1g
…(i)
From Eq. (i) and Eq. (ii), T ′ − 50 = 5a 2 or
T
a
A
and
ms −1
T
19.
B
T'
(c) Net work done on astronaut
×2
T
T T
T
= mghcos mgh cos π = − 72 × 9.8 × 15 = − 10584 Nm
1058.4
(u = 0 ms −1)
− 50 = 50 a
= 11642.4 Nm 11642.4 Nm (b) Work done done on gravitational force
and m2 = 4 kg) = 1 kg and
v2
Now,
F
v =
= ( m1 + m2 ) a
g m1 + m2 = 1 g ( m1 1+ 4 = g = 2 ms −2 5
T
∴
117
T′
= 100 + 10a = 150 N
118 21.
| Mechanics-1
NR cos (180°
− α ) + mgR = 1 mv s2 2
( mg cos α) R ( − cos α) 1 mv s2 2
or
1 × 10 10 × 1 = µ k
+ mgR = 1 mv s2
or
= mgR (1 − cos2 α) 24.
f = force of friction while disc in slipping over inclined surface = µ mg cos cos 30°
N
α
2
10 = 80 µ k + 0.045 10 − 0.045 µ k = = 0.124 80
or
2
1 10 × 2 + × 1 × (0.3)2 × 4 × 10
N
f
s
α
R
( s a y )
s/2
mg
30°
2R
Stops
f' 0.50 m
α or
v s2
or
v s
f ′ = force of friction while disc is slipping over plane surface = µ mg
= 2 gR sin2 α = 2 gR sin α
Now, decreases in PE of disc = Work done against frictional force s mg = fs + f ′ (0.5) 2 s or mg = (µ mg cos 30° ) s + µ mg (0.5) 2
(v s is the speed with which sphere hits ground) 1 1 mv2w = mgR − mv2s 2 2 = mgR − 1 m 2 gR sin2 α 2
∴
vw
or
⇒
= mgR (1 − sin2 α) = mgR cos2 α = 2 gR cos α
(vw is the speed of wedge when the sphere hits ground) 22.
For 45 kg mass to drop 12 mm, the increase in length of the spring will be 24 mm.
Work performed by frictional forces over the whole distance = − mg s = − 50 × 10 × 0.2027 2 1000 2
Now, decrease in PE of 45 kg mass
= Increase in KE of 45 kg mass + Increase in PE of spring 1 45 × 9.8 × 12 × 10−3 = × 45 × v2 2 2 2 × [(75 + 24) − 75 ] × 10−6 i.e., i.e., or or or
5. 5.292 = 22.5 v
2
+ 1 × 1050 2
25.
= − 0.051 J 0.051 J m A g sin θ − µ m A g cos θ − 2 T = m A a T T
+ 2.192
= 3.0996 v2 = 0.13776 v = 0.371 0.371 ms −1
Decrease Decrease in PE of 1 kg mass
= Work done against friction due to 4 kg mass + Increase in KE of 1 kg mass
T
T
22.5 v2
(b) With friction (when mechanical energy does not remain conserved) 23.
mgs (0.5 − µ cos 30° ) = µ mg (0.5) µ × 0.5 s = 0.5 − µ cos 30° 0.15 × 0.5 = = 0.2027 m 0.2027 m 3 0.5 − 0.15 × 2
T a
2T
2T
θ
s o c
cos θ µm Ag cos
A
g m A
S
3
T B mBg
θ
4
or 300 ×
3 5
− 0.2 × 300 ×
4 5
− 2 T = 30a
a
Work, Energy and Power | 180 − 48
− 2T = 30a 132 − 2T = 30a T − m B g = mB a T − 50 = 5a 2T − 100 = 10a
or or Also, or or
m2 g …(i)
28.
32 = 40a a = 0.8 0.8 ms −2
= 2as v = 2 × 0.8 × 1 v = 1.12 1.12 ms −1 2
Work done by frictional force acting on block
∴ Increase in thermal energy of block-floor system
= 66.88 J As the block stopped after traversing 7.8 m on rough floor the maximum kinetic energy of the block would be 66.88 J (just before entering the rough surface). Maximum PE of spring
∴
xmax
= Maximum KE of block = 66.88 =
2 × 66 .88 88 640
0.457 m = 0.457 m Decrease in PE of mass m2 = Work done against friction by mass m1 + Increase KE of mass m1 + Increase in KE of mass m2 10 kg
dU
m1
A
+ ive
B
+ ive
C D
− ive − ive
− ive − ive + ive + ive
E
zero
zero
(U being being lowest − ive) 29.
U =
3
x 3
− 4 x + 6
For U to be maximum (for unstable equilibrium) and minimum (for stable equilibrium) dU =0 dx d x 3 i.e., − 4 x + 6 = 0 dx 3 or
− 4=0 x = ± 2 2 d U d 4 = ( x − 4) = 2x 2 x2
dx
At x At x =
dx
+ 2 m,
d2U dx2
= 2 × ( + 2) = + 4
is minium. ∴U is At x At x = − 2 m,
d2U dx2
m2 = 5 kg
F
x = 6 m point is of stable equilibrium
or
m
Maximum compression compression in the spring 27.
v = 4 ms −1
(b) x (b) x = 2 m point is of unstable equilibrium being + ive) (U being
= − µ mgs = − 0.25 × 3.5 × 9.8 × 7.8 = − 66.88 J 66.88 J
1 2 kxmax 2
2
dr
v
26.
1 1 0 × 4 + (10 + 5) v2 = 0.2 × 10 × 10
Point
2
or
2
Three types of Equilibrium dU (a) F (a) F = − dr
Speed (v) of block A A after it moves 1 m down the plane or
2
Solving,
Adding Eq. (i) and Eq. Eq. (ii), or
× 4 = µ m1 g × 4 + 1 m1v2 + 1 m2v22
5 × 10 10 × 4
…(ii)
119
= 2 × ( − 2) = − 4
is maximum. ∴U is ∴ x = + 2 m point of stable equilibrium. x = − 2 m point of unstable equilibrium.
120 30.
| Mechanics-1
F = −
dU dx
∫ F dx
U=−
i.e.,
Thus, the equilibrium of the charge − q is unstable if unstable if it is slightly displaced along x-axis. x-axis. (b) If char charge ge − q is displaced slightly along Y axis, Y axis, the net force on it will be along origin O and the particle will return to its original position. And as such the equilibrium of the − q is stable.
under F - x graph) = − (Area under F The corresponding U Vs graph will be as U Vs x x graph shown in figure B
F
F'
– q
F'
+q A
B (– a, 0, 0)
E
C
x 32.
D x
–
O
A
(a) Velocity at t = 0 s is 0 ms −1 Velocity at t = 2 s is 8 ms −1 (using v = 0 + at) ∴v = 4 m s −1 (as acceleration is constant)
U +
+q F net
av
= F × vav = ma vav = 1 × 4 × 4 = 16 W at t = 4 s is 16 ms −1 (using
Pav Thus, point C corresponds to stable equilibrium and points A A and E correspond to unstable equilibrium. 31.
(b) Velocity v = u + at)
− q
charge placed at origin is in equilibrium as is two equal and opposite forces act on it. y – q
+q B (– a, 0, 0)
F
+q F
A (+ a, 0, 0)
x 33.
∴ Instantaneous power of the net force at t = 4 s will be P = mav = 1 × 4 × 16 = 64 W Power = Fmin vmax max
(a) But, if we displace it slightly say towards +ive x +ive x side, side, the force on it due to charge to B to B will will decreases while that due to A to A will will increase. – q O
F 2
r
Initially
q F1 > F2
Due to net force on − q towards right the change − q will never come back to original O, its origin position.
F min = (r (r )
F max r
Finally
P = r vmax
∴
vmax
=
P r
Objective Questions (Level 1) Single Correct Option 1.
In KE =
1
mv2
2 m is always +ive and v2 is +ive.
(even if v is − ive).
∴ KE is always + ive.
Below reference level (PE = 0) PE is − ive.
∴ Mechanical energy which is sum of KE and PE may be − ive. ∴ Correct option is (a).
Work, Energy and Power | 2.
Yes, this is work-energy work-energy theo theorem.
=
∴ Correct option is (a).
3.
4.
5.
On a body placed on a rough surface if an external force is applied the static body does not move then the work done by frictional force will be zero. → → W = F ⋅ r21 → → → = F ⋅ ( r2 − r1)
9.
= 50 − 30 + 120 = 140 W 7.
2
4 = 4 ⋅ t 4 0 = 16 J
8.
∴ Correct option is (c). Range = 4 × height 2 2 u2 sin 2θ or = 4 ⋅ u sin θ g
sin 2θ = 2 sin2 θ
or
2 sin θ cos θ − 2 sin2 θ = 0
Work done in displacing the body
or
∴ Correct option is (b). 1 mgh + mv2 W 2 P= = t
= =
t 800 kg × 10 ms −2 1 2
× 10 m +
× 800 kg × (20 ms −1)2 1 min
(800 × 10 × 10) + [400 × (20)2 ] 60
2 g
or or
= 15 J
= 2t
= ma = 2 × 2 t = 4 t W = F v = 4 t × t2 = 4 t3 ∴ Work done by force in first two seconds t =2 = ∫ t = 0 4 ⋅ t3 dt
∴ Correct option is (c). = Area under the curve 1 × 10 = (1 × 10) + (1 × 5) + (1 × − 5) + 2
= t2
F
10.
= (10 ^i + 10 ^j + 20 k^ ) ⋅ (5 ^i − 3 ^j + 6 k^ )
60
Option (c) is correct. t3 x= 3 dx v= ∴ dt dv and a= dt
= [7 x − x2 + x3 ] 50 = 135 J 6.
80000 + 160000
= 4000 W 4000 W
= ( ^i + 2 ^j + 3 k^ ) ⋅ [( ^i − ^j + 2 k^ ) − ( ^i + ^j + k^ )] = ( ^i + 2 ^j + 3 k^ ) ⋅ ( − 2 ^j + k^ ) =− 4+3 = − 1J ∴ Correct option is (b). ∴ W = ∫ F dx 5 = ∫ 0 7 − 2 x + 3 x2 dx ∴ Correct option is (d). → → P = F ⋅ v
121
sin θ (cos θ − sin θ)
=0
cos θ − sin θ = 0 (as sin θ ≠ 0)
or
tan θ = 1
i.e.,
θ = 45°
Now, K Now, K = KE at highest point = 1 m (u cos θ)2 2 = 1 mu2 cos2 θ 2 = 1 mu2 cos2 45° 2 = 1 mu2 ⋅ 1 2 2 1 mu2 = 2 K ∴ 2 i.e., Initial i.e., Initial KE = 2 K
∴ Correct option is (b).
122 11.
| Mechanics-1
P = 3 t2 i.e.,
− 2t + 1 dW = 3 t2 − 2t + 1
15.
dt
dW W
= (3 t2 − 2 t + 1) dt 4 = ∫ 2 (3 t2 − 2t + 1) dt
x = 2.0 m to 3.5 m dU = dx
∴
F = =
12.
t3 2t2 = 3 ⋅ − + t 3 2 2 = [ t3 − t2 + t] 42 = ( 4 3 − 42 + 4) − (23 − 22 + 2) = 52 − 6 = 46 ∆ K = 46 J
∴ x = 4.5
∴
30
= ∫ 20 − 0.1 x dx
KE of 12 kg mass : KE of 6 kg mass = 1 m12v2 : 1 m6v2 2 2
F = = 0 N
= ( 4 × 1.5) + ( − 2 × 1) + 0 = 4 J 1 mv2 = 4 2
1 2
16.
KE at highest point cos 45° )2 = 1 m (u co 2 2 1 2 1 = mu 2 2 = 1 1 mu2 2 2 = E 2
∴ Correct option is (a). 17.
Work done by person
= − [Work done by gravitational pull on rope + =
= m12 : m6 = 12 : 6 = 2 :1
=
= 1 k [ x22 − x12 ] 2
18.
= 1 × 5 × 103 10 − 5 2 100 100 5 × 103 (100 − 25) = 18.75 Nm = 2 × 104 2
∴ Correct option is (c).
× 1 × v2 = 4 v = 2 2 ms −1
[Acceleration [Acceleration being same (equal to g) g) both will have same velocities]
W
− 2 N
∴
30
14.
+4N
m to 5.0 m dU =0 dx
or
x2 = − 0.1 2 20 = − 0.05 × [(30)2 − (20)2 ] = − 25 J 25 J Final kinetic energy = K i + W = 500 + ( − 25) = 475 J 475 J ∴ Correct option is (a).
2
=−4
Work done
Work done by retarding force,
13.
F = =
∴
Correct option is (b). 1 K i = × 10 × 102 = 500 J 2 W
6 1.5
x = 3.5 m 3.5 m to 4.5 m dU 2 = =2 dx 1
4
or
−
gravitational pull on bucket] − mg h + ( − Mgh) 2 M + m gh 2
∴ Correct option is (a). 1 mv2 = Fx 2
∴ x ′ x
=
v ′2 v2
=
1 mv′2 = Fx ′ 2 (2 v)2 = 4 ⇒ x′ = 4 x v2
∴ Correct option is (b).
…(i) …(ii)
Work, Energy and Power | 19.
Vertical velocity velocity (initial) = v0 sin θ
K
At an altitude h
= (v0 sin θ) + 2 ( − (horizontal velocity)2 = (v0 cos θ)2 (Net velocity)2 = v20 − 2 gh Net velocity = v20 − 2 gh (vertical velocity)
20.
2
2
φt 02 g) h
t 0
O
t
Fig. 1
Maximum power will be at 2 s velocity (at t = 2 second) = a ⋅ 2 = F ⋅ 2 m 2F 2 F 2 ∴ Power (at t = 2 s) = F × = m m
21.
123
While ball goes up after elastic collision with the surface it strikes a =
∴ Correct option is (d). v = 0 + at = at = F ⋅ t
H u = gt –0
m
Instantaneous power,
or or
P = Fv = F F ⋅ t m 3 F P = ⋅t m
v = ( − gt0 ) + ( + g) ( t − t0 )
∴
v = − g ( t − 2t0 ) 1 KE = mg2 ( t − 2t0 )2 2
i.e., i.e.,
P = constant t.
K
= φ ( t − 2t0 )2
K
∴ Correct option is (b). 22.
v = u + at
Using,
φt 02
While ball comes down u = 0 a = + g
+
H
O
t 0
2t 0
t
Fig. 2
From the expression for KE
v = = gt 0
= t0 when ball strikes the surface Using v = u + at v = 0 + gt i.e., v = gt. 1 1 KE = mv2 = mg2 t2 ∴
at
at t
2
where,
φ=
2
K = φ t2 1 mg2 = constant. 2
t = 2t0 , K = 0
as shown in Fig. 2. 23.
∴ Correct option is (b). ρ (block) = 3000 kg m −3 = 1000 kgm −3 v σg
v ρg
and
σ
F
(ρ
− )g
(water)
124
| Mechanics-1
σ = 1 ρ 3
i.e., i.e.,
3 y + αx = 5 y + x =1 5 / 3 5 / α
(External force applied to move the block upward with constant velocity).
From above figure
Work done = F s
tan θ =
= v (ρ − σ) gs = v (ρ − σ) × 10 × 3 ( vρ = 5) = 5 (ρ − σ) × 10 × 3 ρ = 5 1 − σ × 10 × 3 ρ = 5 1 − 1 × 10 × 3 = 100 J 3 24.
=
25.
mgH cos π
i.e., i.e., tan φ = tan θ or
Force Constant
/3 l /3 (a)
28. 28.
= 4 t ^i dt 2
2l /3 /3 (b)
^
^
^
0
2
2
0
Given Straight line along which work is zero. → s → F = 20i + 15 j θ
5 α
→ ds
∫ = ∫ (3 t i + 5 j) ⋅ 4 t i dt = ∫ 12 t dt 2
t 3 = 12 3 0 = [ 4 t3 ] 20 = 32 J
Y
5/3
∴ Correct option is (a). → ^ ^ s = 2 t2 i − 5 j → → W = F ⋅ ds
∴ ∴ Correct option is (c).
φ
= α α= 4
Let x Let x be be the elongation in the spring.
∴
1 W ( b) = k′ ( x)2 2 W ( a) > W ( b)
26.
4 3 4
Increase in PE of spring = Decrease in PE of block 1 2 k x = mgx sin θ ∴ 2 2mg sin θ ⇒ x = k
l
k'
=3
⇒ ∴ Correct option is (d).
T
2k'
5 / α 5 / 3 3
or
+ mgH c os 0
H = = Maximum height attained by the projectile = (0) = zero T 1 W ( a) = 2 k′ ( x)2 2
X
4
φ=θ
27.
Time of flight ( T )
=3
Work done will be zero, if
∴ Correct option is (a). Net Net work work done done pav =
15 20
29.
∴ Correct option is (b). W = mgh + mgd = mg ( h + d) or Fav d = mg ( h + d) h ∴ Fav = mg 1 + d
Work, Energy and Power | 30. Energy stored in A = E
= 1 k A x2A
34.
125
Decrease in PE of mass m
2
k A
k B d
A
B
2k 2k B A = 2k F
F
k (spring) (spring) = 100 N/m
F
A
θ
F
B 2
= =
F 1 k A 2 k A F 2
or
2 k A
Energy stored in B in B =
F 2 2 k B
=
F 2 4 k A
=
31.
1 F 2 2 2 k A
( k B
35.
=
E 2
0.5 50
0.15 m = 0.15 m
∴ Correct option is (a). 1 mv2 = F (2x) 1 m (2 v)2 2
…(i)
= F ( nx)
i.e., i.e.,
or
v2
37.
=
2 × 10 × 0.15 −
v = 0.866m .866m s −1
∴ Correct option is (b).
Work done by normal force is zero, being perpendicular to the displacement.
∴ Correct option is (c). Work done on floor = 0 (as displacement is zero).
38.
Acceleration Acceleration
=
(0 − 20) ms −1
=−
k 2 h m
2 gh −
PE of block will change into its KE and then the KE gained by the block will change into the PE of the spring.
=0
2
=
= 4 m.
Net work done = ( + mgh) + ( − mgh)
n = 8 ⇒ ∴ Correct option is (d). 1 1 mgh = mv2 + kx2 2
2 × 10 × 10 × sin 30°
Work done by string is + mgh mgh while that due to gravity it is − mgh. mgh.
…(ii)
Dividing Eq. (i) by Eq. (ii), (2 v)2 n = 2 v2
33.
100 × 22
∴ Correct option is (b). 36.
2
kx 2mg sin θ
Due to inertia the spring will not start compressing the moment the block just touches the spring and as such the block will still be in the process as increasing its KE. Thus v of the block will be maximum when it compresses the spring by some amount.
m k
= 1.5
d=
2
∴ Correct option is (c).
= 2 kA )
2
x = v
2
=
∴ Correct option is (a). 1 2 1 kx = mv2 2
32.
i.e., i.e.,
= Increase in PE of spring 1 mgd sin θ = kx2
10 0.1
× (0.15)2
(10 − 0) s 2 ms −2
Therefore, net force on particle = 2 kg × − 2 m s −2
=− 4N i.e., i.e., the net force on the particle is opposite to the direction of motion.
∴ Correct option is (a).
126
| Mechanics-1
Net work done = F
×s
∴ Correct option is (c).
v-t graph) = F × (Area under v-t graph) = − 4 × 1 × 20 × 10 = − 400 J
KE just before bounce KE just after bounce
2
39.
= 1 mv2 = mgh 2
= 80% of mgh
Thus net work done may not be wholly due to frictional force only.
When the ball attains maxmum height after bounce
Further net force − 4 N may not be wholly due to friction only.
Gain in PE = mgh′ = Loss of KE
Height of bounce
mgh′ = 80% of 80% of mgh
or
= (100 − 20) % of 10 m =8
h′ = 80% of h = 80 % of 10 m
or
=8m
JEE Corner Assertion and Reason Reason 1.
P = Fv
4.
For power to be constant, the velocity must also be constant. constant. Thus, assertion is false. According to 2nd low of motion, net constant force will always produce a constant acceleration. acceleration. Reason is true.
Reason is true.
∴ Correct option is (d). 5.
∴ Correct option is (d). 2.
3.
In circular motion only the work done by The centripetal force is zero. Assertion is false, centripetal force acts towards centre while the velocity acts tangentially.
As displacement is opposite to force (reason) the work done by force will be negative.
As the speed is increasing (slope of graph being increasing) there must be net force in the +ive direction of displacement.
Thus, Assertion is true. Further, as reason is the correct explanation of the Assertion.
Thus, work done to all forces will be positive, Assertion is true and also as explained above reason besides being true is the correct explanation for the Assertion.
∴ Correct option is (a).
∴ Correct option is (a).
Conservative force has nothing to do with kinetic energy. (If a non-conservative non-conservative force acts on a particles, there would be loss of KE). Thus, assertion is false. Work done by conservative force decreases PE (reason is true).
6.
→ Work done by constant force F when the body shifts from A to A to B B.. F
θ
S
A
B
→
→
S'
A
→ F
→
mg (a conservative force)
→ F
S'
→ F
→
S
h
D C
B mg
W
= mgh
If PE at A is A is zero. The PE at B at B would would be − mgh. mgh.
∴ Correct option is (d).
→ → W AB = F ⋅ s = F s cos θ → → Similarly, W AC = F ⋅ s
= F s′ cos (90° + θ) = − F s′ sin θ → → WCD = F ⋅ s = F s cos θ
Work, Energy and Power | → →
greater than 90° and that between N (normal force on wedge) and its displacement will be less than 90° as given in reason.
= F ⋅ s cos(90° − θ) = F s′ sin θ W AC + WCD + WDB = F s cos θ = W AB W DB
∴ Work
done by a constant force is path independent. thus, Assertion is true.
N A
The Reason is false. Kinetic frictional force remains constant but is a non-conservative non-conservative force.
N B
∴ Correct option is (c). 7.
It is true that work-energy theorem can be applied to non-inertial frame also as explained in the answer to question no. 2 of introductory exercise 6.2. Earth is non-inertial which is also true is a separate issue and has nothing to do with the assertion.
For this the work done by N by N (or block) will be negative and that by N (on (on wedge) will be positive as given in assertion.
Thus, option (b) would be the answer. 8.
When block is depressed the excess of upthrust force will act as restoring force and will bring the block up. The velocity gained by block will take the block above its equilibrium and the block will oscillate about its equilibrium position (as given in reason). Thus, the block will be in equilibrium in the vertical direction. Thus, assertion is also true and the reason being correct explanation of the assertion.
∴ 9.
Correct option is (a).
As displacement ( = s2 − s1) is not equal to zero, the work done by all forces may not be zero. Therefore, assertion is false. Work done by all the forces is equal to change in KE is as per work-energy theorem. Thus, reason is true.
∴ Correct option is (d). 10.
127
When the block comes down the wedge, the wedge will move towards left and the actually displacement of normal force will not be along the wedge (see chapter on CM). It will along AB as AB as shown in figure. Thus, the angle between N (normal (normal force on block) and its displacement AB will AB will be
Reason also being the correct explanation of the assertion. 11.
There will be increase in length of the elastic cord. elastic cord
θ
Block Plank f 2
f 1
f 1
s
f
Work-done by static force f 1 (on block)
= − f1s
Work done by static force f 1 (on plank)
= + f1s
∴ Work
done by static force f 1 on the system (block + plank + cord) = 0 Thus, reason is true. The work done by the force F force F will will be used up in doing work against friction f 2 and also increasing the elastic potential energy of the cord. Thus, assertion is true. Further, as the reason is the correct explanation of the assertion.
128
| Mechanics-1
12.
Decrea Decrease se in in KE
= 1 mv2 − 0
PE + KE
2
Rough
In this case the mechanical energy is being used up in doing work against friction and in increasing the PE of the block. ∴ Change ME = 1 mv2 − mgh 2
Case-I
g h m =
s o p P E t o S
h
v'
2
Decrease in ME is used up in doing work against friction.
Slope
f
= 1 mv2
Thus, assertion is true.
30°
As explained above above the reason is false. false.
f'
=
1 2
(µ does not change with the increase in angle of inclination)
2
mv
∴ Correct option is (c).
Change in mechanical energy Objective Questions (Level 2) Single Correct Option 1.
Increase in KE of bead = Work done by gravity + work done by force F force F 1 ∴ mv2 = mgh + FR 2
∴
(Displacement of force F is R is R)) 1 = × 10 × 5 + 5 × 5 = 50 2 100 u= m 14.14 4 ms −1 = 200 = 14.1
3.
Loss of PE of block = Gain in PE of spring d k A
mg ( d + 2) sin 30° or
10 × 10 × ( d + 2) ×
∴ Correct option is (a). 2.
2
2 1
= × 100 × 22 2
d = 2 m Let, v A = velocity of block when it just touches spring
P = Fv
= mav = m v dv v a = dx
dv dt
=
dv ds ⋅ ds dt
= v dv ds
d 2m
m 2 v dv p m 2v 2 ds = v dv p v
∴
ds =
i.e.,
∫
or
1
= 1 k × 22
m
∫
s =
=
m v 3
30°
∴
2v
p 3 v m [(2v) 3 3 p
− (v) 3 ] = 7 mv
∴ Correct option is (a).
A
3 p
3
1 2 mv A = mgd sin 30° 2 2 v A = 10 × 2 × 2 × 1 2 1 − v = 20 ms A
Correct option is (a).
Work, Energy and Power | 4.
Mass per unit length of chain dm = R dθ yCM
=
m π R / 2
129
a2 = 0 ∴ ∴ Correct option is (b).
m π R / 2
6.
Let x Let x be be the expansion in the spring. A
y dm = ∫ ∫ dm
B T
T
θ
C CM
(
2 R 1–π
y
R (1 (1 – cos θ)
π / 2
=
∫
0
R (1 − cos θ)
M 1
)
Increase in PE of spring = Decrease in PE of block C 1 2 kx = M1 gx 2
m R dθ π R / 2
m π / 2
= 2 R ∫ 0 (1 − cos θ) dθ π = 2 R π − 1 π 2 = R 1 − 2 π 1 Now, mv2 + mg yCM 2
or
For block A block A to to remain at rest kx = µ min Mg
∴ 7.
= 2 g y COM 2 v = 2 gR 1 − π v2
mg F (= (= mg )
F = = kx = = mg
8.
F = = mg B mg (weight) (weight)
= 2mg a1 = 2 g ∴ Net force on B on B = 0 Net force on A on A
∴ Correct option is (c). Ti = mg 2mg = mg KX i = When one spring is cut. It means KX i becomes zero. Downward acceleration, acceleration, KX i mg g = = a= 2m 2m 2
A
mg F
M
2
The moment string is cut T T = = mg
= µ min Mg µ min = 2 M 1
2 M1 g
or
∴ Correct option is (c). 5.
kx = 2M1 g
i.e.,
mg
(downward)
Now drawing FBD of lower mass : mg mg − T f = m. a = 2 mg T ∴ f = 2 ∆T = T f − T i = mg or 2 mg sin θ − µ mg cos θ a= m
= g sin θ − µ g cos θ For v to be maximum a dv =0 mg sin θ dx dv or v =0 dx θ
x θ
o s
g c µ m
130
| Mechanics-1
dx dv ⋅ =0 dt dx dv =0 dt
or or
⇒
sin θ = µ cos θ 3 = 3 x ⋅ 4 5 10 5 10 = 2.5 m x = 2.5 m 4
or
⇒
Power delivered by man
13.
= T v cos θ → |T|= T → and |v|= T φ = 3 x + 4 y
(a) Between points E points E and and F F ,
→ →
= T⋅v
→
T
→ v
m
y 3N
∴ Correct option is (d). 9.
x
12.
i.e., i.e., g sin θ = µ g cos θ or
4 µ mg
∴ Correct option is (c).
a = 0
or
x =
dU is − ive. dr
P (6m, (6m, 8m)
4N F net net = 5N
dU , the force between E between E and and F F dr will be + ive i.e., i.e., repulsive. Now, F Now, F = −
10.
F x =
and
→ v → u
PR
→ mg
θ
⇒
→ mg
→
→
PR =
PQ 5 4
×
=5 4
PQ = 10 m
∴ Work done by the conservative force on
→
= m g ⋅ ( u + g t)
the particle → → m g ⋅ g t
= Fnet × PR = 5 N × 10 m = 50 Nm 50 Nm = 50 J 50 J
= mg u cos (90° + θ) + i.e., P = − mg u sin θ + mg2 t → → [|u|= u, |g |= g] Therefore, the graph between P and P and t will be as shown in option (c). 11.
x 2 x
µmg PE of spring due to its compression by x by x
= | Work done by frictional force when displaced by 2 x| x| 1 2 i.e., i.e., kx 2
x
∂ φ = − 3 N − ∂φ ∂ x ∂φ ∂ φ = − 4 N F y = − ∂ y
∴
∴ Correct option is (c). → → Power = F ⋅ v
Q
6m
R
(b) At point C the potential energy is minimum. Thus, C is point of stable equilibrium.
= µ mg 2x
14.
∴ Correct option is (c). Both at x = x1 and x = x2 the
force acting on the body is zero i.e., i.e., it is in equilibrium.
Now, if the body (when at x at x = x1) is moved towards right (i.e. (i.e.,, x > x1) the force acting on it is + ive i.e., i.e., the body will not come back and if the body (when at x = x2 ) is moved toward rght (i.e., (i.e., x > x2 ) the force acting on it is − ive i.e., i.e., the body will return back. Then, x x = x2 is the the position position of stable equilibrium.
∴ Correct option is (b).
Work, Energy and Power | 15.
6b x −7
i.e.,
The man will stop when
131
= 12a x −12 1
x
θ
mg sin sin
i.e., i.e.,
cos θ µmg cos
θ
or 16.
∴ Correct option is (a). 19.
µ mg cos θ = mg sin θ (µ 0 x) mg cos θ = mg sin θ tan θ x = µ0
or
2a x = 6 b
For the rotational equilibrium of the rod
/2 l /2 mg
Taking moment about A about A mg⋅
k1x
kx
l
k2x
C
l = k x l 2 mg
i.e.,
x=
∴ PE
1 stored in the spring spring = kx2 2
2 k
2
A
B
=
F = = (k (k 1 + k 2) x x
AC ⋅ F
∴
AC
=
( k2 x) l
k2 k1
+ k2
∴ Work done by the force of friction = 1 mv2 − mgh 2
10 × 1) = 1 × 1 × 22 − (1 × 10 2
= − 8J ∴ Correct option is (c). a b U = 12 − 6 x
U = ax −12
x
− bx − 6
dU For stable equilibrium, =0 dx i.e., a ( − 12) x −13 − b( −6) x −7 = 0
8 k
∴ Correct option is (c). m − m2 a= 1 g m1 + m2
=
l
∴ Correct option is (d). mgh + work done by the force of friction = 1 mv2
2
= ( mg)
Speed (v (v) with which mass m1 strikes the floor
+ k2 ) x
2
18.
20.
F ( k2 x) l
( k1
= 17.
=
= ( k2 x) l
1 mg k 2 2 k
21.
02
+ 2 gh =
m1 − m2 gh m1 + m2
2
∴ Correct option is (a). F = − ax + bx b x2 ∴ − dU = − ax + bx2 dx
or
dU = ( ax − bx2 ) dx
or
U = ( ax − bx2 ) dx ax2 bx 3 − +c U = 2 3
∫
or
At x At x = 0, F = = 0
∴U = = 0 and so, c = 0 U =
Thus,
ax2 2 3a 2b
U = whe n = 0, when i.e., at i.e., at x x =
=
ax2 2 bx 3 3
−
bx 3 3
132
| Mechanics-1
dU when F = = 0, when F = 0 dx i.e., at i.e., at
= mg ( h + xmax )
…(i)
∴ From above Eq. (i),
− ax + bx2 = 0 a x =
i.e., i.e.,
1 2 k xmax 2
∴
xmax depends upon h and also xmax depends upon k upon k..
b
Graph between U and x and x will will be
KE of the block will be maximum when it is just at the point of touching the plank and at this moment there would no compression in the spring. Maximum KE of block = mgh
22.
3a 2b
a b
O
X 25.
chain
∴ Correct option is (c). W A Fs 1 = = W B
Fs
∴ Correct option is (c). Gain in KE of chain = Decrease in PE of
CM (i (i )
1
2r
∴ Correct option is (c). W A W B
=
1 2 mv A 2 1 2 4 mv B 2
=1 2 v A
⇒
2 v B
v A
⇒
v B
W A W B
=4 1 2 1
=
=1 1 K A
∴ 23.
π
K B
/2 πr /2
=1
πr
1
CM (f (f )
U i at (1, 1) = k (1 + 1) = 2k U f at (2, 3) = k (2 + 3 ) = 5k W
∴ 24.
= U f − U i = 5 k − 2k = 3 k
When the whole chain has justcome out of the tube. 1 2π πr or mv2 = mg + π 2 2
Correct option is (b).
Gain in PE of spring = Loss of PE of block
∴
h
v + 2 gr 2
π +
π 2
∴ Correct option is (b). x max PE = 0 level for block
26.
Acceleration of the block will decrease as the block moves to the right and spring expands the velocity (v (v) of block will be maximum, when 1 1 mv2 = kx2 2 2
Work, Energy and Power | At this moment, F moment, F i.e.,
= kx F x=
29.
k
Work done on block A in A in ground frame (50 − 30) = 0.2 × 45 × 10 × 100
2
2
v
or
∴
27.
=
k F F 2 = m k km F v= mk
= 18 J 30.
∴ Correct option is (b). N = mg cos θ N
∴ Correct option is (b). Force on each block = k x
φ = 53°
4ms –1 A
kx
1 10
= 20 N
= 10 × 10 × cos 37 ° = 10 × 10 × 4 5
= 80 N in 2 s ∴ Work done by N in N cos φ = N cos = 80 × 20 × cos 53 ° = 80 × 20 × 3
i.e., rate i.e., rate of energy transfer = 200 Js 200 Js −1 Correct option is (c).
From O to x to x compression compression in the spring kx
A m kx
B 3m
+ ive
Average acceleration acceleration of A of A kx − F a A = 2m Average acceleration acceleration of B of B kx a B = 2 (3 m)
5
= 960 Nm = 960 J 31.
∴ Correct option is (b). T ′ + mg = 5 m
Upper spring cut
Initially
As at maximum compression of the spring both the blocks would be having same velocity. 2 a A x = 2 aB x [using v2 i.e., i.e.,
a A kx − F 2m
or or i.e.,
→
s = | s |
θ = 37°
= 80 W Power of B of B = 20 × 6 = 120 W ∴ Total power = 200 W
F
→ s
10 ms –1
B kx
Power of A of A = 20 × 4
28.
20 m
6ms –1
= 200 ×
∴
133
6m
kx kx − F = 3 2 kx F = 3 3 F x= 2 k
∴ Correct option is (c).
Lower spring cut
T = T' + + mg + ive
= u2 + 2as]
= aB k x =
(given)
T' mg
T
T' mg
T′
+ m × 10 = 5 m i.e., i.e., T ′ = − 5 m mg − T a= m mg − ( T ′ + mg) = m T ′ =− m
mg
134
| Mechanics-1
=−
( − 5 m)
33.
m = 5 ms −2 32.
Total ME = − 40 J
∴
PE (max)
U ( ( x x )
∴ Correct option is (b). Total ME = 25 J ∴ PE (U ) max = 25 J 25 J [as KE can’t be −ive.]
50 J 25
U ( ( x x )
– 10
50 J PE (max)
– 10
10
10
15
x (m) (m)
– 35 PE (max)
x (m) (m) 6
–5
6
–5
PE (max) 25
= − 40 J 40 J
PE (max)
Particle can't be found in the regions above PE (max) line.
15
∴ “It is not possible”. Option (d).
Particle can’t be found in the region above PE (max) line.
∴ −10 < x < − 5 and 0 < x < 15 ∴ Correct option is (a). More than One Correct Options 1.
→ → (iii) As a ⋅ v
(i) Acceleration U = 7 x + 24 y ∂U = − 7 F x = − ∂ x
∴
i.e.,
∴ i.e.,
a x =
Options (c) and (d) are incorrect. 3i + 4 j v = 3i
– 7i 7i
5 24 a y = − 5 → 7 ^ 24 ^ a =− i− j m/s2 5 5 → |a |= 5 ms −2
= 3 ^i + 4 ^j ∴ ∴
→
−7
∴ Correct option is (b). (ii) Velocity at t = 4 s → → → v=u+at = ( 8.6 ^i + 23.2 ^j ) + − 7 ^i − 24 ^j 4 5 5 = 8.6 ^i + 23.2 ^j − 5.6 ^i − 10.2 ^j → |v| = 5 ms −1
Correct option is (a).
the path of the particle
can’t be a circle. → → i.e., i.e., a is not perpendicular to v.
∂ U F = − 24 y = − ∂ y i.e.,
≠ 0,
→ a 2.
– 24 j
U = 100 − 5x + 100 x2 ∴ F x = − ∂ U = − ∂ (100 − 5 x + 100x2 ) ∂ x ∂x
∴
= − [ − 5 + 200 x] = 5 − 200 x 5 − 200 x F a x = x = 0.1
0.1 = 50 − 2000 x (i) At 0.05 m from origin x = + 0.05 m a x = 50 − 2000 (0.05)
= 50 − 100
(first point)
Work, Energy and Power |
= − 50 m/s2 ive x = 50 m/s2 towards − ive x
Work done by spring = Energy stored in the spring = 1 kx2 2
∴ Correct option is (a). (ii) At 0.05 m from origin x =
− 0.05 m
∴ Correct option is (b).
(second point)
(iii) If spring is initially its natural length and finally compressed. 1 Work done on (not by) the spring = kx2 2 will be stored in the spring.
a x = 50 − 2000 ( − 0.05)
= 50 + 100 = 150 m/s2 ∴ Correct option is (c).
∴ Option (c) is incorrect.
Mean Position 50 a = 0 at x at x = m = 0.025 m 2000
(iv) If spring is initially at its natural length and finally extended. Work done on (not by) the spring = 1 kx2 2 will be stored in the spring.
(iii) For point 0.05 m from mean position x = (0.05 + 0.025) m
= 0.075 m ∴ a x ( at x = 0.075 m ) = 50 − 2000 (0.075) = 50 − 150 = − 100 m/s2 100 m/s2 towards − ve x ve x-axis. -axis. = 100 m/s ∴ Correct option is (b). (iv) For second point 0.05 m from mean position x = 0.025 m 0.025 m
− 0.05 m 0.05 m
= − 0.025 m ∴ a x = 50 − 2000 ( − 0.025) = 100 ms −2 incorrect. ∴ Option (d) is incorrect. 3.
(i) If the spring is compressed by x by x,, elastic 1 2 potential energy equal to kx gets stored 2 in the spring. Now, if the compressed spring is released the energy stored in the spring will be lost. When the spring attains to natural length. Work done by spring = Energy stored in the spring = 1 kx2 2
∴ Option is (d) 4.
is incorrect. incorrect.
(i) Work-Energy theorem states that W net (Work done by all forces conservative or non-conservative, non-conservative, external or internal)
= ∆ (KE) Correct option is (d) is incorrect.
∴ Correct option is (c). (ii) Work done done by non-conservative forces (i.e., i.e., all forces except conservative forces) lead to decrease in KE and thus change in mechanical energy takes place.
∴ Correct option is (b). (iii) Work done by by a conservative conservative force may may be + ve, − ve or zero. s
90° mg h
h
mg
mg (a mg (a conservative force)
∴ Correct option is (a). (ii) If the spring is extended by x, x, energy 1 2 stored in the spring would be kx . If the 2 extended spring is released the energy stored in the spring will be lost when the spring attains its natural length.
135
W
(PE increases)
→ →
= m g⋅h mgh cos π = mghcos = − mgh
136
| Mechanics-1
W
→ →
When F is removed, the upper disc accelerates upwards and when it attains the position as in figure 2, its acceleration acceleration reduces to zero and the velocity gained by it takes it further upwards. Restoring force on the upper plate now acts downwards and that on the lower plate acts in the upward direction and would lift it (lower plate) if
= m g⋅h = mghcos0 mgh cos0 = + mgh
(PE increases) W
π = m g ⋅ h = mghcos mgh cos = 0 → →
2
(PE remains same)
k (δ − l) > 3 mg
Option (a) is incorrect. in correct. 5.
i.e.,
is the force by hand or upper disc F is
or
(δ – l )
or
k ( (l + + δ)
k ( (δ + l )
6.
3mg k l l = = mg l
7.
mg
At maximum extension extension x x : decrease in potential energy increase in spring energy 1 ∴ (2m)( g)( x) = kx2 2 4 mg or k= k
k (l + + δ)
Artist to make it 3/2 times of l
of B =
Total work done by internal forces of a system, which constitute action and reaction pairs, is always zero and if it is not so the total work done will not zero.
∴ Correct options are (b) and (c). 8.
Uncomprssed/ Unstretched spring
k l l 3mg
N
k
Correct option is (a) obviously being incorrect.
N
3mg
+ kl kδ > 3 mg + mg δ > 4 mg
∴ Correct option is (d).
F k ( (δ + l )
kδ > 3 mg
∴ Option (a) is incorrect. Correct option is (b).
= 3 mg + k ( l + δ) = 3 mg + k l + 2mg
Correct option is (c). (ii) In pure rolling work done by frictional force (a non-conservative force) is always zero.
k
= 3 mg + kl + 2mg = 3 mg + mg + 2mg = 6mg for δ = 2mg k
∴ Correct option is (b). N = kl + 3 mg = mg + 3 mg = 4 mg ∴ Correct option is (c).
(i) Work done by conservative forces may be + ive, − ive or zero as explained in the answer to question no. 4
∴ Correct option is (d). 9.
In moving from 1 to 2 work done by conservative force
= U1 − U 2 = ( − 20) − ( − 10) = − 10 J 10 J ∴ Option (a) is incorrect.
Work, Energy and Power |
137
Option (b) is correct.
Work Work done done by Norm Normal al forc force e ( N on N ) on block A block A
Work done by all forces
= Ns cos θ In motion 2 → + ve In motion 1 → + ve ∴ Correct option is (b). W
= ( K 2 + U2 ) − ( K1 + U 1) = [20 + ( − 10)] − [10 + ( − 20)] = 20 J ∴ Correct option is (c). N
In motion 1 → may be − ive if f is f is directed upwards along the plane as shown in figure 1. (Motion 1 being retarded)
N 2
1
mg
as θ < 90°
Work done by force of friction ( f ( f ))
Option (d) is incorrect. 10.
as θ < 90°
∴ Correct option is (c). In motion 1 → may be + ive if f is f is directed
mg
downwards along the plane if motion 1 is accelerated.
Work done by gravity In motion 2 → − iv ive
π In motion 1 → zero[as θ = ]
[as θ = π]
∴ Correct option is (d).
2
∴ Correct option is (a). Match the Columns Columns 1.
As body is displaced displaced from x from x = 4 m to x = 2 m → (a) F
(d) → (p, s). 2.
→ s
(a) → (p)
= 4 ^i
∴
(b) Work done by mg will mg will be − ive as θ =
= 4 i − 2i ^
= Fs cos θ
(a) Work done by N will will be + ive as θ < 90°
= − 2 ^i m
→ → W = F ⋅ s
W
^
= − 8 unit
N
(a) → (q) and| and |W |= 8 unit (a) → (s) → ^ ^ (b) F = 4 i − 4 j → → ∴ W = F ⋅ s = ( 4 ^i − 4 ^j) ⋅ ( − 2 ^i) = − 8 unit
(b) → (q) (c) Work done by force of friction ( f ( f )) will be zero as f will f will be zero. (c) → (r)
→ → W = F ⋅ s
(d) Work done by by tension tension (T (T ) will be + ive as θ < 90°
^
= 8 i unit
and |W |= 8 unit (c) → (p, s) → ^ ^ (d) F = − 4 i − 4 j → → ∴ W = F ⋅ s
and
T
mg
(b) → (q, s) → ^ (c) F = − 4 i
∴
π
= ( − 4 ^i − 4 ^j) ⋅ ( − 2 ^i) = 8 unit |W |= 8 unit
(d) → (p) Y
3.
+q A (– a, 0)
F B
F A = F B +Q F net = 0
+q B (+ a, 0)
at A = force on + Q by + q placed at A F at B B = force on + Q by + q placed at B
F A
138
| Mechanics-1
As F A = F B charge equilibrium.
+ Q
will
be
in
4.
(a) From A From A to to B B : :
(a) A
F A'
FB'
+q
+q A
B
+Q
< F A
Due to decrease in distance between (at B (at B ) ) and + Q F B′
+q
kx
Mean position
Due to increase in distance between + q (at A) A) and+ Q F A′
x
Increase in spring PE
B mg
= mgx = ( kx) x = kx2 (a) → (q)
∴
Thus, equilibrium will be a stable one.
Increase in KE of block
(b) From A From A to to B B
= Decrease in gravitational PE of block − Increase in spring PE = kx2 − 1 kx2
Y F net
A
F A +Q B
+q
2
Decrease in gravitational. PE of block
> F B
F B
+q
As F net will be along the increasing direction of Y , Y , the charge + Q will not return to origin.
2 1 2 = kx < Decrease in gravitational PE of 2 block.
∴
(b) → (p)
(c) From B From B to C
(c) As explained in (b) the equilibrium will be an unstable on.
1 Increase in spring PE = k (2x)2 2 3 2 = kx 2 1 Decrease in KE of block = kx2 2
∴
(KE of block at C will be zero)
Thus, equilibrium will be an unstable one.
∴
(b) → (q)
(c) → (q)
(d)
x = y line line
F net
− 1 kx2 2
∴ (c) → (p). (d) From B From B to to C decrease in gravitational PE = mgx
F A
= ( kx) x = kx2
A +q
C
= 1 kx2
Using F Using F A = F B , we have F have F B′ > F A . As there will net force on + Q which will being + Q to origin. (b)
x
+q
3 Increase in spring PE = kx2 2
∴ (d) → (p)
Work, Energy and Power | 5.
a=
2−1 − m2 g = g m1 + m2 2+1 m1
139
(c) Work done by string on 2 kg block
= Ts cos π = 40 × 0.15 × ( −1) 3
= − 20 J
a
∴ (c) → (s).
T
T
(d) Work done by string on 1 kg block
T
m2
= Ts cos0 = 40 × 0.15 × 1
a
T
3
m1 1g
=
1 3
∴ (d) → (q).
g
6.
3 2 m1m2
(a) Work done by friction force ( f )
(w.r.t. ground)
= 10 m s −2 T =
= 2J
2g
f
g
+ m2 1×2 =2× × 10 1+2 m1
f' (= (= f )
= f s cos π = − f s
= 40 N 3
Displacement of blocks, 1 s = ut + at2 2 1 2 = at 2 = 1 × 10 × (0.3)2 2 3
= 0.15 m 0.15 m (a) Work done by gravity on 2 kg block mgs cos0 = mgscos0 = 2 × 10 × 0.15 × (1) =3J
∴ (a) →
(r)
(b) Work done by gravity on 1 kg block mgs cos π = mgscos = 1 × 10 × 0.15 × ( − 1) = − 1.5 J 1.5 J
∴ (b) → (p).
∴ (a) → (q). (b) Work done by friction force on incline (w.r.t. ground)
= f × 0 × cos 0 =0 [There being no displacement of incline w.r.t. ground]
∴ (b) → (r). (c) Work done by a man in lifting a bucket
= Ts cos0 (T = = Tension in rope) = a + ive quantity and s both s both would be in upward θ = 0 as T and direction.
∴ (c) → (p). (d) Total work done by friction friction force in (a (a) w.r.t. ground
= − f s + 0 = − ∴ (d) → (q).
fs
