PHYSICS SOLUTIONS ADVANCE ADVANCE LEVEL PROBLEMS PR OBLEMS TARGET : JEE (IITs)
TOPIC : WORK, POWER & ENERGY PART - I 1.
W=
(4t dt dt i ) ! F .ds = ! (3t i " 5 j) . (4t ˆ
ˆ
2
=
! 12 t
2
dt
=
ˆ
0
2.
2
# $
12 t 3 3
0
= 32 J
dW = 0 ; dx % x = ! , x = 0 i. e. F(x) = 0 Clearly for d = ! ,the work done is maximum.
For W to be maximum ;
Alternate Solution : External force and displacement are in the same direction & Work will be positive continuously so it will will be maximum when displacement is maximum. 3.
Work done in changing speed from 0 to V is W 1 =
1 mV2 2
work done in changing the speed from V to 2V is W 2 =
& 4.
1 m (2V)2 2
–
1 1 mV2 = 3 mV2 2 2
W1 1 W2 = 3
Applying Applying work energy theorem on block
W F + W S = 0 F! –
1 k !2 = 0 & 2
work done 5.
! =
2F k
or
2F2 = F! = k
In the frame (inertial ( inertial w.r.t w.r.t earth) of free end of spring, the initial velocity of block is 3 m/s to left and the spring unstretched .
WORK, POWER & ENERGY - 1
Applying conservation of energy between initial and maximum extension state. 1 1 mv2 = kA2 2 2 6.
7.
or
m v = k
A=
Work done : = Mgh1 + Mgh2 + Mgh3 + µ1 Mg !1 + µ 2 Mg !2 + µ3Mg !3 = Mg (h1 + h 2 + h 3) + Mg (µ 1!1 + µ2!2 + µ3!3) = Mg (8 + 0.2 + 0.4 + 0.4) = 90 J From conservation of energy K.E. + P.E. = E
&
&
or K.E. = E –
- 2E * 1 ( = 0 k + , k ) 2
After the top end of chain falls down by
2g
!
8
=
g! 2
8
, the speed of chain is
7 M. 8
momentum of chain is 7 M 8
9.
!
.
The mass of chain above table is
&
2E is zero. k
The speed of particle at x = –
v=
1 kx2 2
2E is k
K.E. at x = –
E –
8.
4 ' 3 = 6cm. 10000
g! 2
=
7 M 16
g!
Let v be the speed of B at lowermost position, the speed of A at lowermost position is 2v. From conservation of energy 1 1 m (2v)2 + mv2 = mg (2!) + mg!. 2 2 Solving we get v =
10.
6 g! 5
Let h be the height of water surface, finally a 2h = a .
a a a . ; h= 2 2 4
- a a * " ( = a – 3a = 5a 8 8 , 4 8 )
&
C.M. gets lowered by a – +
&
Work done by gravity = mg
5a 8 WORK, POWER & ENERGY - 2
11.
It can be observed that power delivered to particle by force F is P = Fv = K. The power is constant. Hence work done by force in time t is .W = Pt = Kt
12.
The work done by force from time t = 0 to t = t sec. is given by shaded area in graph below. Hence as t increases, this area increases.
& 13.
Work done by force k eeps on increasing.
Power
/ /
P = F . V = FV
- dm * ( , dt )
F=V +
d(6 ' volume 2 =V5 4 1 dt 3 0
6 = density
5 d( volume 2 1 dt 3 0
= 6V 4
= 6V (AV)
= 6AV2
& Power P = 6AV3 or P 8 V 3 14.
P=
d (mgh) dt
1000 ' 10 ' 100 50 Pact = 2000 W Pact =
Pconsumption = 15.
2000 W = 16 kW.. 0.25
When 4 coaches (m each) are attached with engine (2m) according to question P = K 6mgv ..............(1) (constant power), (K being proportionality constant) Since resistive force is proportional to weight Now if 12 coaches are attached P = K.14mg.v1 ............(2) Since engine power is constant So by equation (1) and (2)
%
6Kmgv = 14Kmgv1 6 × 20 14 = v1 = 8.5 m/sec Similarly for 6 coaches =
%
v2 =
6 ×20 8
v1 =
6 ×v 14
=
6 ' 10 60 = 7 7
%
K6mgv = K8mgv2
=
3 × 20 = 15 m/sec 4 WORK, POWER & ENERGY - 3
16.
Let F the force with which man pulls the block. & & Fv = 500 F = 50 N (F - mg) v = 100 solving m = 4 kg
17.
At point 'C', the potential energy is minimum, hence it is a point of stable equilibrium. Also, from E to F, the slope is negative i.e.,
%
F = –
dU 90 dr
F is +ve so repulsive
Hence, the force of interaction between the particles is repulsive between points E and F. t
PART - II 1.
(a) Assume 20 kg and 30 kg block to move together 50 = 1 m/s2 50
&
a=
&
frictional force on 20 kg block is f = 20 × 1 = 20 N
The maximum value of frictional force is
fmax =
1 × 200 = 100 N 2
Hence no slipping is occurring. The value of frictional force is f = 20 N. & Distance travelled in t = 2 seconds. 1 × 1 × 4 = 2m. 2 Work done by frictional force on upper block is W fri = 20 × 2 = 40 J Ans. Work done by frictional force on lower block is = – 20 × 2 = – 40 J Ans. (b) Yes Ans. (c) Wor k done by frictional force on the upper block is converted to its kinetic energy. Ans. S=
2.
- 3 L0 * k + ( 2 , 4 ) 1
%
2
v=
=
k m
1 2
mv 2 +
1 2
k (L 0
: x) 2
when x < L0
= - 3 L * 2 @ 2 0 ?+ ( : ; L 0 : x< B ?> , 4 ) BA 2
when x
C L0
1 - 3L 0 * 1 K+ ( = mv 2 2 , 4 ) 2
%
v=
3L 0 4
k m
which is also the maximum speed of the block. T hus, vmax =
3.
(a)
3 L0
k
4
m
For motion to start 5 D k mg > Dsmg 4
or
5Dk > 4Ds
WORK, POWER & ENERGY - 4
(b)
At the final position of the block extension in spring is maximum and the speed of the block is v = 0. Hence the net work done in taking the block from initial to final position .W = work done by P + work done by spring force F + work done by friction = .K = 0 x
!
= P x – Kx 3 . dx – µkmgx = 0
5 D k mg Kx 4 x – 4 4
–
Dkmgx = 0
1 / 3
- DK mg * ( solving we get x = + , K ) 4.
At the moment, elongation is maximum, speed of the block becomes zero. Applying W/E theorem on system 1 . mv2 = W g + W s 2
0 – 1 2
: . 2.4 = 2.10 . x –
1 . 100 . x2 2
- 3 " 1 * + ( Solving x = + 5 ( m , ) 5.
Applying work-ener gy theorem between A and B.
3 m 37° 3m
1 1 mV B2 – mV A2 = W gravity + W friction 2 2
%
1 1 mV B2 – m (136) = – mg(3 + 3 sin 37°) – µmg cos 37º x 3 2 2
VB2 % 2
–
136 = – 48 – 12 2
% V B = 4 m/s
WORK, POWER & ENERGY - 5
6.
Maximum chance of slipping occurs when spring is maximum compressed. At this moment, as force exerted by the spring is maximum, acceleration of the system is maximum. Hence maximum f riction force is required at this moment. By W/E theorem 1 1 (M + m) V2 = Kxm2 2 2 Now for upper block
am =
%
xm =
(M " m) V 2 K
kx m M"m
force on upper block is provided by the friction force. Therefore Dmg > V = Dg
For limiting value
kx m .m M"m
M"m k
using values Vmaximum = 20 cm/s
7.
The speed is maximum when acceleration is least Let displacement of block is x0 when the speed of block is maximum. At equilibrium, applying Newton’s law to the block along the incline mg sin E = µmg cos E + kx0 ..................(1) Applying work energy theorem to block between initial and final position is Kf = Ki + mg x0 sin E –
1 kx 02 – µ mg x 0 cos E ...........(2) 2
Solving (1) and (2) we get,
8.
Vmax = (sinE – µ cos E) g
m k
Ans.
Vmax = (sin E – µcosE) g
m K
(i)
Velocity will be maximum when net force = 0.
k.x = DN = D mg
%
x =
D mg k
W s + W f = .K 2
2
Dmg 1 2 1 - 2Dmg * 1 - Dmg * k+ ( : k+ ( : Dmg ' % = mv 2 , k ) 2 , k ) k 2 On solving v = Dg
m . k WORK, POWER & ENERGY - 6
(ii)
When the particle have velocity equal to zero, then let extension in spring be ' x '.
- 2 D mg * k + ( 2 , k ) 1
2
D2
2
2 2
m g
= 2
k
- 2 D mg * 1 2 " x( + k x , k ) 2
= D mg +
D2
2 2
m g
+ x D mg +
k
x = 0 (at natural length) or x = :
1 2
= 2 D mg @ " x B = 0 > k A 2 D mg
%
k x 2
x?
2 D mg
when compression in spring is i.e. initially k k So at natural length, velocity is zero and spring force is also zero. The block will not return or have velocity towards left. 9.
(a)
The particle is at equilibrium when F = 0
%
x (3x – 2) = 0
%
x = 0 and x =
2 m 3
(b)
The particle is in stable equilibrium at x = 0 m etre and unstable equilibrium at x =
Since at
(c)
x=0
dF F0 dx
%
d 2U dx 2
2 x= m 3
9 0 and at
dF 90 dx
%
The minimum speed imparted to the particle should be such that it just reaches x =
2 metre 3 d 2U dx 2
F0
2 from there on 3
it shall automatically reach x = 0 1 mv2 = – 2
10.
F = –
GU ˆ Gx i
–
2 / 3
2 / 3
! F dx = ! x (3x : 2) dx = 1300 27 –
4
v=
:4
2600 m/s 27
GU ˆ ˆ ˆ j jˆ jˆ Gy = – [6 i ] + [8] = – 6 i + 8 "
"
& a = – 3 ˆi + 4 jˆ has same direction as that of u H "
or
: 3ˆi " 4 jˆ 2
"
- a * H ++ (( , 2 )
"
|a| = 5
| u | = 5/2
"
"
Since u and a are in same direction, particle will move along a straight line
& S = 11.
5 1 ×2+ × 5 × 22 2 2
= 5 + 10 = 15 m.
According to W.E. theorem 5 1 mV 2 - 0 = (10 " 4 x ) dx 0 2 V = 10m/s Force at that moment = (10 + 20) = 30 N Instantaneous power = 30 × 10 = 300W
!
WORK, POWER & ENERGY - 7
