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RSM79-P1-WEP-PH-17

6. Solved Problems 6.1 Subjective

Problem 1: A particle of mass m is displaced from a position P1 to P2 with position vectors kcjbiar1 ˆˆˆ

and kbjaicr2

ˆˆˆ

by a force

kajcibF ˆˆˆ

. Find the work done by the force.

Solution: If the displacement of the particle is S

, the work done W by the given force F

is equal to F

. S

, when F

is a constant force. S.FW

…(1)

where 12 rrrS

)]kcjbia()kbjaic[(S

]k)cb(j)ba(i)ac[(S

…(2) and kajcibF

…(3)

Using (1), (2) and (3) ]k)cb(j)ba(i)ac).[(kajcib(W

W = (c-a)b + (a-b)c + (b-c)a

W = bc - ab + ac - bc + ab - ac = 0

Net work done by the force F

for the given displacement is zero. Problem 2 : A block is projected horizontally on a

rough horizontal floor. The coefficient of friction between the block and the floor is . The block strikes a light spring of stiffness k with a velocity v0. Find the maximum compression of the spring.

v0

k

m

Solution: Since the block slides and the spring is compressed through a

distance x the net retarding force acting on it = F = kx + N = ( mg + kx)

Work done by net force for the displacement x, W = dx.F

W = x

0dxF

KE= x

0dxkxmg

m

x

2

kxxmg0mv21 2

20

x2 + kmg2 x- 2

0vkm = 0

N

mg

kx

N


RSM79-P1-WEP-PH-18

x = 2

kmv4

kgm4

kmg2 2

02

222

x =

1

gv

mk1

kmg

20 .

Problem 3: Two smooth balls of mass m1 and m2

connected by a light inextensible string are at the opposite points of horizontal diameter of a smooth semi cylindrical surface of radius R. If m1 is released, find its speed at any angular distance moved by m2.

m2

R

m1 m2

Solution: Let the ball m2 move through an angle . The mass m will fall through

a distance h1 = R. The ball m2 rises through a height h2 as, h2 = R sin . The change in gravitational potential energy of m1 is PE1 = -m1gh1 = -m1 gR (since m1 loses its potential energy as it falls down). The change in gravitational potential

energy of m2 is PE2 = m2gh2 = m2gR sin (since m2 gains potential energy as it rises

up) The total change in gravitational potential energy = PE = PE1 + PE2

m2

R

m1

v m1

h1

h2

v

PE = -m1gR + m2gR sin = gR (m2 sin - m1). The change in KE of the system (m1 + m2)

= KE = 21 m1v2 +

21 m2v2 =

2v)mm( 2

21 …(2)

where v = speed of m1 and m2 at the positions as shown in the figure.

From the principle of conservation of energy we obtain, KE + PE = 0 …(3)

Using (1), (2) and (3), we obtain,

21 (m1 + m2)v2 – gR (m1 – m2 sin ) = 0

v = 21

21

mmsinmmgR2

.


RSM79-P1-WEP-PH-19

Problem 4 : A block of mass M initially has a velocity v0 when it just touches a spring. The block moves through a distance l before it stops after compressing the spring. The spring constant is k and the coefficient of kinetic friction between block and table is . As the block moves the distance I, (a) what is the work done on it by the spring force? Are there other forces acting on the block, and if so, what work do they do? (b) what is the total work done on the block? (c) use the work-energy theorem to find the value of l in terms of M, v0, , g and k.

Solution: (a) The net force acting on the block by

the spring is equal to Fspring = kx Work done by the spring

= ds.Fspring

= Fspring ds cos 180 = Fspring.ds

=2kl2

l

0

dxkx

mv0

mv

kxmg

k

(b) The total work done = W = KE = 0 (1/2) m 2

0v = (1/2)m 20v .

(c) The work done by friction = - mgl The total work done =-mgl – (1/2)kl2

= - 20mv

21 2

02 mv

21mglkl

21

l

1g

vmk1

kμmg

20

Problem 5 : A wedge of mass M with a smooth quarter circular

plane, is kept on a rough horizontal surface. A particle of mass m is released from rest from the top of the wedge as shown in the figure. When the particle slides along the quarter circular plane, it exerts a force on the wedge. If the wedge begins to slide when the particle exerts a maximum horizontal force on it, find the coefficient of friction between the wedge & the horizontal surface.

M

m

Solution : In order to find the coefficient of

friction at the time of maximum horizontal thrust (Fx)max exerted on the wedge by the particle given by the formula

= N

fmax . . . (i)

Fx

mg

v

a

h

O

Mg

N

f

1

2

R


RSM79-P1-WEP-PH-20

Where fmax = maximum static friction between the wedge & ground. (limiting friction ) that must be equal to the maximum horizontal force (Fx)max for prevalence of sliding of the wedge & N = normal force offered by the horizontal surface on the wedge.

(F)max & N can be calculated as follows. Let the particle attain a speed v at the angular position as shown in the free body diagram.

Since the particle accelerates towards the centre O with an

acceleration a = r

v 2

, the force exerted on it must be radially inwards.

R – mg sin = ma = r

mv 2

R = mg sin + r

mv 2

. . . (ii)

Conserving energy of the particle between position 1 & 2 we obtain

(KE)12 = (PE) 12 21 mv2 = mgh = mr sin . G

v = singr2 . . . (iii)

Elimination of v between (ii) & (iii) yields R = 3 mg sin . .. (iv)

the horizontal force acting on the wedge = Fx =R cos

Fx =(3 mg sin ) cos = 3mg sin cos = 23 mg sin 2 .. . (v)

For Fx to be maximum sin 2 = 1 =450 Putting in equation (v) horizontally

we obtain ( Fx)max = 23 mg . . . (vi)

Resolving forces acting on the wedge for its equilibrium along horizontal & vertical we obtain, fmax – (Fx)max = Ma = 0 & N – Mg –R sin = May = 0

fmax = (Fx)max = 23 mg & N = Mg + R sin | = /4 =

m

23M g

using (I) the values of fmax & N we obtain

= m3M2

m3

.

Problem 6: In the figure shown stiffness of the spring is k and

mass of the block is m. The pulley is fixed. Initially the block m is held such that, the elongation in the spring is zero and then released from rest. Find :

(a) the maximum elongation in the spring (b) the maximum speed of the block m. Neglect the mass of the spring, pulley and that

of the string.

m


RSM79-P1-WEP-PH-21

Solution: (a) Let the maximum elongation in the spring be x, when the block is at position 2.

The displacement of the block m is also x. If E1 and E2 are the energies of the system when the block is at position 1 and 2 respectively. Then

E1 = U1g + U1s + T1 Where U1g = gravitational P.E. with respect to surface S.

U1S = P.E. stored in the spring. T1 = initial K.E. of the block.

2

1m

S

h1

h2m

E1 = mgh1 + 0 + 0 = mgh1 . . . (1) and E2 = U2g + U2s + T2

= mgh2 + 21 kx2 + 0 . . . (2)

From conservation of energy E1 = E2

mgh1 = mgh2 + 21 kx2

mgxhhmgkx21

212

x = 2mg/k

(b) The speed of the block will be maximum when it is at the

equilibrium point. Let xo be the elongation in the spring. When the block is at equilibrium point

from conservation of energy

mgh1 = mg(h1 - xo) + 2mv21 + 2

0kx21

mg2

222

kgmk

21mv

21

kmg

mg =kxo

2

mvkgm 222

v = g km .

Problem 7: A heavy particle hanging from a fixed point by a light inextensible

string of length l is projected horizontally with speed lg . Find the speed of the particle and the inclination of the string to the vertical at the instant of the motion when the tension in the string is equal to the weight of the particle.

Solution : Let tension in the string become equal to the

weight of the particle when particle reaches the point B and deflection of the string from vertical is . Resolving mg along the string and perpendicular to the string, we get net radial force on the particle at B i.e.

FR = T - mg cos . . . (i)

B

Amgsin

mgcos

OT


RSM79-P1-WEP-PH-22

If v be the speed of the particle at B, then

FR = l

2mv . . . (ii)

From (i) and (ii), we get

T - mg cos = l

2mv . . . (iii)

Since at B, T = mg

mg(1 – cos ) = l

2mv

v2 = g (1 - cos) . . . . (iv) Conserving the energy of the particle at point A and B, we have

22o mv

21cos1mgmv

21

l

where vo = gl and v = cos1gl

g = 2g(1 - cos) + g (1 - cos) cos = 2/3 . . . . (v) Putting the value of cos in equation (iv) we get

v = 3gl

Problem 8 : A block of mass m starts from rest and slides

down the surface of a frictionless solid sphere of radius R as shown in figure. Measure angles from the vertical and potential energy from the top. Find

(a) the change in potential energy of the mass with angle

m

R

(b) the kinetic energy as a function of angle (c) the radial and tangential acceleration as a function of . (d) the angle at which the mass flies off the sphere. Solution: (a) Consider the mass when it is at the

point B. UA (P.E. at A) = 0 UB(P.E. at B) = -mgR (1 – cos ) U = UB - UA U = - mgR(1 - cos) Negative sign indicates that P.E.

decreases as particle slides down.

B

A

R(1 - cos?

R

(b) Conserving energy at points A and B. UA + TA = UB + TB where UA = P.E. at A, UB = P.E. at B

TA = K.E. at A, TB = K.E. at B

0 + 0 = -mgR(1- cos) + TB


RSM79-P1-WEP-PH-23

T = mgR(1 – cos )

(c) Since T = 2mv21

2mv21 = mgR(1 – cos ) = mgR 2sin2 /2

v = gR2 sin/2

aradial = v2/R

ar = 4g sin2/2

at = g sin

also, in circular motion velocity is along the tangent, therefore

vtangential = 2 gR sin /2

at = tvdtd

at = (gR) cos/2 dtd = (gR) cos/2

at = (g/R) v cos/2 , as v = 2 gR2 sin /2 at = gsin (d) For circular motion

mg cos - N = R

mv2

at the moment when the particle breakes off the sphere N = 0.

mg cos = R

mv2

g cos = v2/R

- v = 2 gR sin/2

g cos = 4g sin2/2 = 2g(1 - cos)

cos = 2/3 = cos-1(2/3)

mgcos

mgsin

N A

Problem 9: A small block is projected with a speed

v0 on a horizontal track which turns into a semi circle (vertical) of radius R. Find the min value of Vo so that the body will hit the point A after leaving the track at its highest point. The arrangement is shown in the figure, given that the straight part is rough &

Rough surface

R

smooth v0

A


RSM79-P1-WEP-PH-24

the curved path is smooth. The coefficient of friction is .

Solution : Let the block escape the point at C with a velocity V horizontally. Since it hits the initial spot A after falling through a height 2R we can write (2R) = ½ gt2 where t = time of its fall.

t = 2 g/R

The distance AB = 2v g/R

R

v1

d

v

C

B A

d = 2v g/R . . . (a)

Work energy theorem is applied to the motion of the body from A to B leads ΔKE = W f

dmgVm21vm

21 2

12o

vo = dg2v21 . . . (b)

Energy conservation between B & C yields

)R2(mgmv21mv

21 22

1

v1 = gR4v2 . . . (c)

When the bead escapes C, its minimum speed v can be given as

R

mv 2

= mg (the normal contact force = 0)

v= gR . . . (d)

By using (c) and (d) we obtain v1 = gR5 . . . (e)

using (a) and (d) we obtain

d = R2gR2gR

. . . (f)

Putting the values of v1 and d in (b) we obtain vo = )R2(g2gR5

vo = gR45 μ . Problem 10: A heaped chain of total length is kept

at the edge of the smooth table top. One end of the chain is pulled down slightly & released, the links of the chain far from the heap gradually under the weight of the hanging part of the chain. Find the speed of the


RSM79-P1-WEP-PH-25

hanging chain in the function of the overhanging part.

x

Solution: Suppose the chain acquires a velocity v after falling through a distance x let the corresponding acceleration be a. Since the nearest link of the portion 1 to the portion 2 is expelled from the heap 2 with a relative velocity v, impact force R comes into play on the moving portion 1 of the chain due to the heap. Resolving the forces acting on the portion 1 we obtain,

x

mg – R = ma . . . (i)

where m = mass of the hanging portion of the chain = LM x,

R = -v dtdm = v

v

LM = 2v

LM

substituting the expression for m in equation (i) we obtain,

xLMv

LMgx

LM 2

a

gx – v2 = x (vdxdv ) . . . (ii)

Let v2 = kx

2v dv = k. dx

dxvdv =

2k ( putting

dxvdv =

2k )

& v2 = kx in (ii) we obtain,

)kg(2k

k = 32 g

v2 = (2/3) gx


RSM79-P1-WEP-PH-26

6.2 Objective Problem 1: When a man walks on a horizontal surface with constant velocity, work

done by (A) Friction is zero (B) Contact force is zero (C) Gravity is zero (D) Man is zero Solution : Since mg & N are perpendicular to velocity

v

and sd

, work done by these forces is zero. Since no relative sliding occurs during walking, static friction comes into play. Hence the point of application of static frictional force does not move. Therefore no work is done by frictional force.

f

N

mg

v

ds

The man has to lose his body's (internal) energy E, hence performs work because W = E.

Problem 2: A pumping machine pumps a liquid at a rate of 60 cc per minute at a

pressure of 1.5 atmosphere. The power of the machine is (A) 9 watts (B) 6 watts (C ) 9 kW (D) None of these Solution : Power = F.v, where F = force imparted by the machine , v = velocity

of the liquid P = p.A.v, Where p = pressure & A = effective area

= pdtdV = )1060(10

23 65

( 1 atm 105 N/m2 )

= 9 watts. Problem 3: A particle of mass m is moving horizontally with a constant velocity v

towards a rigid wall that is moving in opposite direction with a constant speed u. Assuming elastic impact between the particle and wall, the work done by the wall in reflecting the particle is equal to:

(A) (1/2) m(u+v)2 (B) (1/2) m(u+v) (C) (1/2) muv (D) None of these Solution: Velocity of separation = - e (velocity

of approach) (-v'i) - (-ui) = -e[(vi) – (-ui] u – v' = -1(v + u) v' = 2u + v

W = KE = 22 mv21'mv

21

= 22 v)vu2(m21

= 2mu (u + v).

u

v

u

v '

y

x


RSM79-P1-WEP-PH-27

Problem 4: A block of mass m moving with a velocity v0 on a smooth horizontal floor collides with a light spring of stiffness k that is rigidly fixed horizontally with a vertical wall. If the maximum force imparted by the spring on the block is F, then:

(A) mF (B) kF

(C) 0vF (D) None of these

Solution : Energy conservation

between the positions A and B PE+ KE = 0

21 kx2 -

21 m 2

0v =0

x = 0vkm

Fmax = kx = 0vkm Fmax k m v0.

m

v0 k

m

B

x

Problem 5: A particle of mass m moves on the x-axis under the influence of a force

of attraction towards the origin O given by F = -k/x2 i . If the particle starts from rest at a distance a from the origin the speed it will attain to reach the origin will be:

(A) 2/1

axxa

mk2

(B)

2/1

axxa

mk2

(C)

xaax

mk (D)

2/1

axxa

k2m

Solution : F = 2xk

2xk

dxvdvm

vdv = 2xdx

mk

x

a2

v

0 xdx

mkvdv

x

a

2

x1

mk

2v

=

a1

x1

mk

v = ax

x)(am2k

.


RSM79-P1-WEP-PH-28

Problem 6 : A block of mass m is moving with a constant acceleration a on a rough horizontal plane. If the coefficient of friction between the block and ground is , the power delivered by the external agent after a time t from the beginning is equal to:

(A) ma2t (B) mgat (C) m(a+g)gt (D ) m(a+g)at Solution : Instantaneous power delivered = P

= v.F

= Fv where, F – f = ma F = f + ma P = (f + ma) v Put f = mg P = (mg + ma)v = m(a + g).at

Ff

am

v

Problem 7: A particle moves with a velocity k6j3i5 m/s under the influence of

a constant force N)k20j10i10(F

. The instantaneous power applied to the particle is:

(A) 200 J/s (B) 40 J/s (C) 140 J/s (D) 170 J/s Solution : P = F

. v

= (5 i – 3 j + 6 k ) .(10 i + 10 j + 20 k ) = 50 – 30 + 120 = 140 J/s

Problem 8: A spring placed horizontally on a rough horizontal surface is

compressed against a block of mass m placed on the surface so as to store maximum energy in the spring. If the coefficient of friction between the block and the surface is , the potential energy stored in the spring is

(A) k2

gm 222 (B) k

gm2 22

(C) k2

gm22 (D) kmg3 22

Solution : For equilibrium of the

block Fmax - mg = 0 F = mg

U = k2

gmk2

F 2222

Fmax

mg

m


RSM79-P1-WEP-PH-29

Problem 9: An object of mass m is tied to a string of length L and a variable horizontal force is applied on it which starts at zero and gradually increases until the string makes an angle with the vertical. Work done by the force F is:

(A) mgL(1-sin) (B) mgL (C) mgL(1-cos) (D )mgL(1+cos) Solution : Work done by the external force

= E of the system (object) = PE ( 0KE ) = mgh = mg (L – L cos ) = mgL (1 – cos ).

h m

B

L

m A

L cos

Problem 10: When a body of mass M slides down an inclined plane of inclination ,

having coefficient of friction through a distance s, the work done against friction is:

(A) Mg cos s (B) Mg sin s (C) Mg ( cos sin )s (D) None of the above Solution : Work done against friction

= - work done by friction = mg cos s.

Problem 11: An electric motor creates a tension of 4500 N in hoisting cable and

reels it at the rate of 2 m/s. What is the power of electric motor? (A) 15 KW (B) 9 KW (C) 225 W (D) 9000 H.P. Solution : P = T.v (numerically)

= 4500 2 = 9000 W = 9 kw Problem 12: A body is acted upon by a force which is inversely proportional to the

distance covered. The work done will be proportional to: (A) s (B) s2 (C) s (D) None of the above Solution : W = sd.F

= dssks

s1

= kln (s/s1)

Problem 13: A body starts from rest with uniform acceleration and acquires a

velocity V in time T. The instantaneous kinetic energy of the body the body in time t is proportional to:

(A) (V/T)t (B) (V2/T)t2 (C) (V2/T2)t (D) (V2/T2)t2


RSM79-P1-WEP-PH-30

Solution : KE = (1/2) mv2

= 2)at(2m

= 2

tTV

2m

TV

tva

K.E. 22

2t

TV

Problem 14 : A body moving at 2m/s can be stopped over a distance x. If its kinetic

energy is doubled, how long will it go before coming to rest, if the retarding force remains unchanged ?

(A) x (B) 2x (C) 4x (D) 8x Solution : If R = resistance

Rx1 =KE1 = KE . .. (i) Rx2 = KE2 =2KE . . . (ii)

(ii) (i) 2xx

1

2

x2 = 2x1 = 2x Problem 15 : A machine delivers power to a body which is proportional to velocity of

the body. If the body starts with a velocity which is almost negligible; then distance covered by the body is proportional to

(A) v (B) 32v

(C) v5/3 (D) v2

Solution : Power = Fv = mdtdv (v) v

mv dtdv = k0v where k0 = constant

m dtdv = k0 mv

dxdv = k0

vdv = dxmk0

integrating both sides ,

x

0

0v

0

dxmkvdv

xmk

2v 0

2

hence, displacement is proportional to square of instantaneous velocity.

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