WORK, ENERGY & POWER IIT-JEE Syllabus 1.
Kinetic and potential energy
2. Work and power
Total No. of questions in Work, Power & Energy are: Solved examples…....…………………………..…20 Exercise # 1 …….……………………………….…25 Exercise # 2 …….……………………………….…37 Exercise # 3 …….……………………………….…28 Exercise # 4 ……………………………………..…07 Exercise # 5 ……………………………………..…06 Total No. of questions………………..123
*** Students are advised to solve the questions of exercises in the same sequence or as directed by the faculty members.
Index : Preparing your own list of Important/Difficult Questions CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000
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Instruction to fill (A) Write down the Question Number you are unable to solve in column A below, by Pen. (B) After discussing the Questions written in column A with faculties, strike off them in the manner so that you can see at the time of Revision also, to solve these questions again. (C) Write down the Question Number you feel are important or good in the column B.
EXERCIS E NO.
COLUMN :A
COLUMN :B
Questions i am unable to solve in first attempt
Good/Important questions
1
2
3
4
5
Advantages 1. It is advised to the students that they should prepare a question bank for the revision as it is very difficult to solve all the questions at the time of revision. 2. Using above index you can prepare and maintain the questions for your revision.
KEY CONCEPT CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000
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(a) Work done by a constant force ( F ) during displacement ( S ) of point of application of force is
velocity v w.r.t S. A block m is acted upon by a constant force F. The block is initially at rest in S - frame. In - S- frame Work done by F in time t W = F.S
1 F 2 F2 t 2 t = = F 0.t 2 m 2m W=
F.S ;
W = Fx(x2 – x1) + Fy(y2 – y1) + Fz(z2 – z1) ˆ Where F = Fx ˆi Fy ˆj Fz k r1 = x1ˆi y1ˆj z1kˆ r2 = x 2 ˆi y 2 ˆj z 2 kˆ S = r2 r1 POSITIVE AND NEGATIVE WORK : (b) The work is said to be positive if the angle between force and displacement is acute (< 90° ) because W=
F .S
= F S cosand cos = (+)ve
In S - frame Work done by F in time t
1 F 2 t W’ = FS’ = F vt 2 m Clearly W W´ (e) LINE INTEGRAL OF FORCE
Work done by a variable force F to move from r1 to r2 r2
when < 90º. W=
F. dr
r1
d r small displacement
The positive work means that the external force supplies energy to the system or body. (c) The work is said to be negative if the angle between force and displacement is obtuse ( > 90°) because cos= (–)ve when > 90°.
When the force changes in three dimensions then we can consider it to be of the form ˆ F = Fx ˆ i + Fy ˆj + Fz k For calculating the work done from A to B while going through the path AB, we calculate it for an infinitesimal displacement d r i.e. B
W=
F.dr
A
The negative work means that the force is extracting energy from the system. (d) WORK IS RELATIVE : Work done by a force depends on the frame of reference. Suppose S & S’ are two frames of reference and S’ is moving with a constant CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000
Where
d r = dx ˆ i + dy
ˆj +
ˆ dz k
WORK, POWER & ENERGY
38
B
W=
(Fxdx + Fydy + Fzdz)
A
Now using path equation, eliminate y and z from Fx, x and z from Fy, x and y from Fz and integrate the above function xB
W =
yB
Fxdx +
xA
Fydy +
yA
Fzdz
zA
(g) WORK DONE BY SPRING FORCE
= (KEf – KEi) + (PEf – PEi)
(a) Conservative Force (i) If the work done by a force in moving a body from an initial location to a final location is independent of the path taken, then the force is conservative. W1 = W2 = W3
1 2 2 k ( x initial x final ) 2
(Elongation/compression) xinitial & xfinal are measured from natural length of spring.
(a) K =
(iv) Wby force other than conservative
zB
(f) Area under force vs displacement graph gives the work done. (Area is considered with proper sign).
Wspring =
= (KEf – KEi)+Wagainst conservative force
(ii) If the total work done by a force is zero, when a body is moved in a closed path, then the force is conservative.
1 mv2 2
m mass of the body
WACBDA = WACB + WBDA = 0
v speed of the body (b) WORK ENERGY THEOREM: (i) The sum of the work done by all the forces [external as well as internal] acting on a particle (or a system of particles) is equal to the change in its kinetic energy. Work done by all the forces W = (K.E.)final – (K.E.)initial (ii) While applying work energy theorem, we must take into account not only the external but also the internal forces to calculate the total work done. (iii) Wcosnservative force + Wnon = K.Ef – K.Ei
conservative force
(iii) For conservative force F = 0 ˆ i
ˆj
ˆ k
x
y
0 z
Fx
Fy
Fz
where = ˆi x ˆj y kˆ z ˆ F Fx ˆi Fy ˆj Fz k
(b) Non-Conservative force (i) The work done by a force in a closed path is non-zero.
Wby forces other than conservative force = (KEf – KEi) – Wby conservative force CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000
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Force is directed along (–)ve x - axis At D,
dU d2U = 0 and <0 dx dx 2
Force = 0 , position D = Unstable equilibrium position
WABCDA 0 (ii) Work done by force is path dependent.
At E,
(c) Potential Energy Curve
if we slightly displace the particle from position E, it experiences no force. Thus position E = neutral equilibrium position.
(i) Conservative force is defined as the negative gradient of the potential energy
(ii) FC = – U = – grad U ˆ ˆ ˆ = i x j y k z
= del
operator U = scalar potential energy function (iii) In one dimension, the component of conservative force along a specified direction (say x -axis ) is equal to the negative derivative of the potential energy w.r.t. distance along the x – axis. i.e. Fc x = –
dU = 0 and dx
U x
(iv) If we plot U(x) versus x curve for one dimensional motion then the negative of the slope is equal to the conservative force.
(a) The change in the potential energy of a system corresponding to a conservative internal force is defined as f
Uf – Ui = – Wc =
Fc . d r
i
where Wc = Work done by the internal conservative force on the system. (b) Gravitational potential energy (Near the earth’s surface)
UB – UA = –WAB = mgh
Fx = – At A,
U x
dU = (–)ve Fx = (+)ve dx
Force is directed along + (ve) x-axis At B,
dU d2U = 0 and >0 dx dx 2
Force = 0, position B = stable equilibrium position
dU At C, = (+)ve , Fx = (–)ve dx CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000
U=
1 2 kx 2
x elongation/compression in spring k spring constant
(a) Mechanical energy = kinetic energy + potential energy E=K+U (b) The total mechanical energy of a system remains constant if WORK, POWER & ENERGY
40
(i) The internal forces are conservative . (ii) and the external forces do no work. This is called the principle of conservation of mechanical energy. Note :– We can apply the principle of conservation of mechanical energy even in the presence of external forces. But in that case, external forces must not work on the system.
(i) The time rate of doing work is called the power delivered by the force. (ii) Instantaneous power
d dW r F . P= = = dt dt Where
F. v
F = applied force
v = instantaneous velocity (iii) Average power
(c) MATHEMATICAL FORM
Pavg=
Uf + Kf = Ui + Ki = E = constant (d) Modified form of the work- energy theorem. According to the work-energy theorem, Wc + Wnc + W ext = Kf – Ki But
.....1
work done by force during given time interval given time int erval
(iv) Area under Power versus time graph gives the work done
Wc = – ( Uf – Ui )
Wnc + Wext = (Kf + Uf) – (Ki+ Ui) Wnc + W ext = Ef – Ei = E .....2 (i) We generally come across with complicated integration technique while calculating the work done by the conservative force. To avoid this complexity, we use this modified form. (ii) While applying this modified form , we must not include the work done by those conservative forces whose corresponding potential energy terms are to be included. (e) If the internal forces are only conservative but the external forces can do work , then Wext = Ef – Ei
[ Wnc = 0]
i.e. the work done by the external forces equals the change in the mechanical energy.
CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000
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SOLVED EXAMPLES Ex. 1
A person pulls a 6 kg block by a distance of s = 12 m along a horizontal surface at a constant speed. If the coefficient of kinetic friction is = 0.2 and the cord pulling the block is at an angle of 45° with the horizontal, then calculate the work done by the person. (take g = 10 m/s2)
Sol.
From simple geometry =
– 4 2
dW
W=
W =
=
F. ds
Fds
cos
4 2
Sol.
=
/ 2
0
=
sin 4 2
=
Work done W = Fs cos
Ex. 2
Ex.3 2
–
RF
2
/ 2 0
2RF sin sin 9 9
0.2 6 10 = 1 = 10 2 newton (1 0.2) 2
= (10 2 ) × 12 ×
R d
=
mg cos sin
1
[ ds = R d]
Since there is no upward motion N + F sin = mg .....(1) Since the motion along the displacement direction is without any acceleration F cos = N .....(2) from eq. (1) & (2) F=
4 2
F cos
= 120 J
Figure shows a smooth circular path of radius R in the vertical plane which subtends an angle of /2 at O. A block of mass m is taken from position A to B under the action of a constant force F which is always directed towards the point B, then determine the work done by this force.
Sol.
4RF 2
= 2 2 RF
A chain of mass m and length rests on a rough–surfaced table so that one of its ends hangs over the edge. The chain starts sliding off the table all by itself provided the over hanging part equals of the chain length. What will be the total work performed by the friction forces acting on the chain by the moment it slides completely off the table? Initially for the chain g = µ(1– ) g µ=
1
...(1)
Let at an arbitrary moment of time, the length of the chain on the table is x.
(mg – F) 20 =
1 mv2 (from work-energy 2
theorem)
Net friction force Ff = N = xg or
Infinitesimal work done by friction
dW = – Ff
. dr = –Ff ds
(50 – F) 20 =
1 × 5 × (10)2 2
or 50 – F = 12.5 or F = 50 – 12.5 dW = – (µxg)(– dx) = F = 37.5 N gx dx Work done by the force 1 = – 37.5 × 20 = – 750 joule Hence the total work done (the negative sign. is used because the push of g the air is upwards while the displacement is W = dW = 1 downwards.) 0
xdx
Ex.5
(1 )
W = – (1– ) Ex.4
Sol.
m mg [= ] 2
figure. The coefficient of friction between the block and the plank is , and its value is such that the block becomes stationary w.r.t the plank before it reaches the other end.
An object of mass 5kg falls from rest through a vertical distance of 20 m and reaches a velocity of 10 m/s. . How much work is done by the push of the air on the object ? (A) 350 J (B) 750 J (C) 200 J (D) 300 J (B) The following two forces are acting on the body (i) Weight mg is acting vertically downward (ii) The push of the air is acting upward. As the body is accelerating down ward, the resultant force is (mg – F). Workdone by the resultant force to fall through a vertical distance of 20m is (mg – F) × 20 joule. 1 mv2 2 Now the work by the resultant force is equal to the change in kinetic energy i.e.
A plank of mass M and length L is placed at rest on a smooth horizontal surface. A small block of mass m is projected with a velocity vo from the left end of it as shown in the
Gain in the kinetic energy =
Sol.
(i) Find the work done by the friction force on the block during the period it slides on the plank. Is the work positive or negative? (ii) Calculate the work done on the plank during the same period. Is the work positive or negative? (iii)Also, determine the net work done by friction. Is it positive or negative ? From Free Body Diagrams :-
For Block a1 =
Ff =g m
W2 =
velocity at time t v1 = v0 – µgt For Plank a2 =
=
Ff mg = M M
velocity at time t, v2 =
mv 0 1 M M m 2
1 2
m 2 Mv 02
m M 2
mg t M
negative In the position 1 (figure) the spring of constant K is undeformed. Find the work
done by the force
(which is always
F
directed along the tangent to the smooth hemi spherical surface) on the small block of mass m to shift it from the position 1 to position 2 slowly.
Finally both moves with common velocity i.e. v1 = v2
t=
= positive
(iii) Net work done by the friction is : mM 1 W = W1 + W2 = – v02 = 2 m M
Ex.6
v0 – µgt =
2
mg t M
Mv 0 ( M m)g
common velocity v =
mv 0 Mm
Sol.
Let us locate the block at an arbitrary angular position < dW = F . dr = Fds = F (Rd) ...(1) Now from the condition of equilibrium of the block , we have F = mg cos + KR ...(2) from eq. (1) & (2) dW = mg R cos d + KR2 d Hence the sought work
(i) Work done by the friction on the block = Change in the kinetic energy of the block W1 = Kf – Ki =
1 1 m v2 – m v02 2 2
mv 1 0 W1= m 2 M m
=–
1 2
mM
2
W =
v 02
m M 2
0
0
(M + 2m ) v02
= negative (ii) Work done by the friction on the plank = Change in the kinetic energy of the plank W2 =
1 Mv2 – 0 2
dW
0
0
cos d KR 2
d
W = mg R sin + Ex.7
=
KR 2 0 2 2
A pendulum of mass m and length is suspended from the ceiling of a trolley which has a constant acceleration a in the horizontal direction as shown in the figure . Find the
maximum deflection of the pendulum from the vertical.
(a) Show that the maximum height reached by the stone is :
v 02 F 2g 1 W
h=
(b) Show that the speed of the stone upon impact with the ground is Sol.
Free Body Diagram w.r.t trolley
WF W F
v = v0 Sol.
(a) From work – energy theorem, For upward motion, work done by (gravity+airdrag)=change in the energy –
Initial velocity of the mass m = 0 Velocity at maximum deflection = 0 K=0 Applying work- energy theorem, we get Wg + Wps + WT = K
h=
+
= ma sin
) a 1 cos = g sin
Ex.8
a g
A stone with weight W is thrown vertically upward in the air with initial speed vo. If a constant force F due to air drag acts on the stone through out its flight :
F 2g 1 W
...(2)
Dividing eq. (2) by (1), we have v2 WF WF v = v0 2 = WF v0 WF
– mg(1– cos) + ma sin + 0 = 0
a = g 2
W 1 W 2 gh – Fh = v –0 g 2 g
h=
( = maximum deflection angle ) ( T
= 2 tan–1
...(1)
v 02
WT = work done by tension = 0
Hence
F 2g 1 W
(b) For downward motion
Wps= work done by pseudo force
tan
W 1 W gh – Fh = 0 – g 2 g
v 02
= – mg (1 – cos )
kinetic
v02
Wg = work done by gravity
ds
1/ 2
Ex.9
1/ 2
The potential energy (in Joule) of a particle of mass 1 kg. moving in the XY plane obeys the law = 3x + 4y, where (x, y) are the coordinates of the particle in metre. If the particle is at rest at (6,4) at time t = 0, then find out. (i) Its acceleration (ii) The work done by the external force in moving the particle from the position of rest of the particle, to the instant of the particle crossing the X - axis
Sol.
(iii)The speed of the particle when it crosses the y – axis. (iv) The co-ordinate of the particle at time t = 4 sec. (i)= 3x + 4y x
ˆ F = – = – i
–
ˆj
y
r
= r0 + u t +
1 (– 3ˆi 4ˆj )t2 2
= ( 6ˆi 4ˆj ) + 0 +
...(1) = xˆi yˆj When the particle is crossing x -axis, y = 0 4 – 2 t12 = 0 hence t1 = 2 sec
Work done W = F . ( r – r0 ) = kf – ki W = ( 3ˆi 4ˆj ) . ( 3ˆi 4ˆj ) × 2 = 9 + 16 = 25 J (iii) When the particle is crossing y –axis, x = 0 hence from eq. ....(1) 3 2 6– t = 0 or t2 = 2 sec. 2 2 =
u
+
3 2 4 2
ydx
W =
1 ˆ ˆ ×( r = ( 6i 4 j ) – 2
4×4 i 28ˆ j = – 18ˆ Coordinates of the particle = ( –18, – 28 )
=a
dW
x2
x1
=
b y 2 2 y12 2
xdx + b
y2
y1
ydy
b x 2 2 x 12 2
+
= Independent of the path
Hence “ F = ax ˆ i + by ˆj ” is potential. Alternate :
For conservative field of force × F = 0
(i) F = ay ˆ i
× F =
F i j k I × (ay ˆ ) G Hx y z JK i
ˆ =– k
× 2 = 10
m/s (iv) Using eq.(1), at t = 4 sec.
3ˆ i 4ˆj )
x1
(ii) W = F . dr = (ax ˆ i + by ˆj ) . (dx ˆ i + dy ˆj ) = ax dx + by dy
t2 v = a t2 =
x2
“ F = ay ˆ i ” is not potential.
Hence
t2 = 0 +
a
=a
dW
Value of integral depends on the type of the y(x)function, that is on the shape of the path.
1 a t2 2
a
W=
position at time t,
m
(m = 1 kg ) (ii) acceleration is constant,
v
F = – 3ˆi 4ˆj m/s2 Acceleration a =
Sol.
(i) F = ay ˆ i (ii) F = ax ˆ i + by ˆj Where a and b are constants .Are these fields potential (conservative ) ? Let us calculate the work performed by the force of each field over the path from a certain point 1 (x1,y1) to a certain point 2 (x2 , y2) (i) dW = F . dr = (ay ˆ i ).(dx ˆ i + dy ˆj ) = ay dx.
ˆ ˆ F = – (3 i + 4 j ) N
r
There are two stationary fields of force
Ex.10
y
ˆ 0 (ay) = –a k
F = ay ˆ i is a non conservative force
(ii) F = ax ˆ i + by ˆj
× F = ˆj
) =0
F I G Hi x j y k z JK× (ax ˆi
+ by
ˆ ˆ F = ax i + by j
is a conservative
(b) F = –
force Ex.11
a
b , where a 2 r r and b are positive constants and r is the distance from the centre of the field. Find (a) The value of r0 corresponding to the
–
–
3
r
b 3a
dr
2
= =
6a r
F r 3a b
–
4
L d UO M M P Ndr P Q=
=0
bI F G H3a JK – 3
= 2a
b
2
=
2b 3 27a 2
–
b3
b3 27 a
2
Ex.12
A uniform rod of mass M and length L is held vertically on a horizontal surface as shown in figure. Assuming zero potential energy at the base of the rod, determine the potential energy of the rod.
Sol.
Consider a small element dy at a height y from the surface; mass of the element = dy
b 2a 2 3 r r
d dr
r3
27 a 2 Maximum magnitude of the attraction force
= 0 r0 =
r02
2b
3a b
=
2a b d2U
+
9a 2
r2
–
3 r0
r
4
b3
b
b
r2
=
At equilibrium position r = r0 , F = 0 2a
b
L O M N P Q
2a
6a
r =
equilibrium position of the particle, examine whether this position is steady (stable) (b) The maximum magnitude of the attraction force dU (a) F = – dr a b d F=– dr r 2 r F=
–
dF =0 dr
The potential energy of a particle in a certain field has the form U =
Sol.
2a dU = dr r3
2b r3
2
1
2
r03
r ro
6a 2b r0
b 2a
3
= 6a 2a .b 2 b
=b
b 2a
3
>0
The position is steady (stable ).
1960 N/M. If the mass of block A is 2 kg, calculate the mass of block B and the energy stored in the spring .
M dy L Potential energy of the element, M dU = (dm) gy = gy dy L dm =
Potential energy of the rod, Mg L dU = U = 0 L
z
ydy
=
MgL 2 Ex.13 A uniform chain is held on a frictionless table with one-fifth of its length hanging over the edge. If the chain has a length m and a mass m, how much work is required to pull the hanging part back on the table? (A) mg/10 (B) mg/5 (C) mg/50 (D) mg/2 Sol.(C) Mass of the hanging part of the chain = (m/5); the weight mg/5 acts at the center of the gravity of the hanging chain, i.e., at a distance = /10 below the surface of a table .
The gain in the potential energy in pulling the hanging part on the table. mg U = × = 5 10 mg 50
Work done = U = mg /50 Ex.14
Two blocks A and B are connected to each other by a string and a spring ; the string passes over a frictionless pulley as shown in figure . Block B slides over the horizontal surface of a stationary block C and the block A slides along the vertical side of C, both with the same uniform speed. The coefficient of friction between the surfaces of the blocks is 0.2 .The force constant of the spring is
Sol.
Free Body Diagram
Since Block A and B has no acceleration and spring is massless. N = m Bg ....(1) T = N = mB g
....(2)
T=T’
....(3)
T = mAg
....(4)
’
From eq.(2), (3) & (4) mBg = mAg mB =
mA
=
2 0.2
= 10
kg . Extension in the spring = x (say) T = kx T 19.6 x= = = 0.01 m k 1960 Energy stored in the spring U=
1 2 kx 2
=
Ex.15
A 2 kg block collides with a horizontal weightless spring of force constant 2 N/m.The block compresses the spring 4 metre from the rest position. The speed of the block at the instant of collision (i) If the surface on which the block slides is frictionless, (ii) if the coefficient of kinetic friction between the block and the horizontal surface is 0.25 . (A) 4 m/s, 4m/s (C) 5m /s,7.96 m/s
Sol.
= 19.6 Joule Now total energy lost by the block = 16 Joule + 19.6 Joule = 35.6 Joule If v1 be the velocity of the spring at the instant
1 × 1960 × (0.01)2 = 0.098 J 2
of collision, we have 1 mv12 = 35.6 Joule 2 1 × 2 × v12 = 35.6 2 v12 = 35.6
(B) 4m/s , 5.96m/s (D) 5 m/s, 5m/s
(B) The situation is shown in fig.
or Ex.16
v1 = 5.96 m/sec.
An ideal massless spring can be compressed 2 metre by a force of 200N. This spring is placed at the bottom of a frictionless inclined plane which makes an angle = 30° with the
(i) When the spring is compressed, the energy E stored is given by 1 2 1 kx = × 2 × 42 = 16 Joule 2 2 This amount of energy is imparted to the block. Let v be the velocity of the block at the instant of collision. When spring is compressed, the velocity of block becomes zero. Hence, the energy lost by the block, is– E=
=
1 1 mv2 – mv02 2 2
1 = mv2 as v0 = 0. 2
horizontal. A 20 kg mass is released from rest at the top of the inclined plane and is brought to rest momentarily after compressing the spring 4 metre. Through what distance does the mass slide before coming to rest ? (A) 4.17 m (B) 1.00 m (C) 8.17 m (D) 2.17 m Sol.(C) As the spring is compressed by 2 metre with the application of a force of 200 N. Hence its force constant k is given by F 200 = = 100 N/ m 2 2 Let be the distance along the inclined plane k=
in which the mass travels before it comes to rest. Now applying the conservation of energy.
1 mv2 = 16 Joule 2
1 × 2 v2 = 16 or v = 4 m/sec. 2 (ii) Work done by the block against friction, W = Force × Distance or
= R. distance
(F = R)
= (mg) d
( R =mg)
= 0.25 × 2 × 9.8 × 4
1 2 kx = mg h = mg sin 2
1 2 or Ex.17
Limiting (maximum ) static frictional force between block and track Ffmax= s mg = 0.22 × 0.5 × 10 = 1.1 N
1 × 100 × 42 = 20 × 9.8 × × 2
or
F < Ff max = 800/98 = 8.17 m.
A 0.5 kg , block slides from the point A on a horizontal track with an initial speed 3m/s towards a weight less horizontal spring of length 1m and force constant 2 N/m . The part AB of the track is frictionless and the part BC has the coefficient of static and kinetic friction as 0.22 and 0.20 respectively. (see figure ) If the distance AB and BD are 2m and 2.14m respectively, find the total distance through which the block moves before it comes to rest completely. (g = 10 m/s2)
Ex.18
The block will no move back So, the total distance moved by the block = 2 + 2.14 + 0.1 = 4.24 m An automobile of mass m accelerates, starting from rest, while the engine supplies constant power P. show that. (a) The velocity is given as a function of time 2Pt m
1/ 2
by v =
(b) The position is given as a function of time 1/ 2
8P 9m
by, s = Sol.
t3/2
(a) Given that power = Fv = P which is constant. mdv v=P dt t P v ovdv = o m dt Pt v2 = m 2
z
Sol.
2Pt m
Kinetic energy of the block 1 1 mv2 = × 0.5 × 32 = 2.25 J 2 2 Path AB is frictionless . In the path BD , work done against friction = K mgs =
= 0.2 × 0.5 × 10 × 2.14 = 2.14 J So at D, kinetic energy = 2.25 – 2.14 = 0.11 J Now if the spring is compressed by x, from energy conservation.
1/ 2
v= (b) v =
ds dt
s
ds
o
2P m
2P m
1/ 2
8P 9m
1/ 2
s=
Ex.19
1 × 2 × x2 + 0.2 × 0.5 × 10x 2 x2 + x – 0.11 = 0 x = 0.1 m [ x = –1.1 is in admissible]
A particle moving in a straight line is acted by a force, which works at a constant power and changes its velocity from u to v in passing over a distance x. The time taken will be -
vu 2 2 v u
2 2 3 v u x 2 v 3 u 3 v u (C)
(C)
Sol.
t 1 / 2 dt
t 3/ 2
(A) x
Compressed spring exerts a force F = kx = 2 × 0.1 = 0.2 N
t
2 3/ 2 t 3
0.11 =
0
.
s=
1 2 0.11 = kx + Kmgx 2
1/ 2
=
vu 2 2 v u
(B) x
(D)
x
The force acting on the particle =
mdv dt
mdv Power of the force = dt (constant)
m at
2
v 2
= kt + c
mu 2 2
v2 mu 2 = kt + 2 2
1 m (v2 – u2 ) = kt 2 mdv Again v=k dt dv m.v v=k dx v
k dx
u
Ex.20
Sol.
...(2)
x
mv 2 dv =
0
1 Intergrating, m (v3 – u3) = kx ...(3) 3 From eqn (2) and (3), 2 2 3 v u t= x 2 v 3 u 3 2 ˆ 2 ˆ Under the force F = xy i + yx j , a particle is moving along a parabolic path given by y = x2. Find the work done by this force in moving the particle from (0, 0) to (a, a2) 2 ˆ 2 ˆ Given F = xy i + yx j But i + dy ˆj dr = dx ˆ
W= =
F . dr
path
xy
2
dx yx 2 dy
path
a2
a
=
xy
x 0
2
dx +
xy
y0
W = =
Now from (1), m
a
0
...(1)
c =
t = 0, v = u
v = k
Eliminating y2 and x consecutively from (i) and (ii) function, we have
2
dy
x 5 dx +
a2
y
2
0
a6 a6 a6 + = 6 3 2
dy
