BEST COACHING INSTITUTE FOR IIT-JEE BOOKLET (WORK, ENERGY AND POWER)

1

WORK, ENERGY AND POWER WORK DONE BY A CONSTANT FORCE Consider a particle, that undergoes a displacement ∆r along a straight line while upon acted by a constant force

F that makes an angle θ with ∆r as shown in figure. F θ

F cos θ

∆r

r r The work ‘W’ done by a constant force F acting on a particle is equal to the product of the component of F along the direction the displacement of particle and the magnitude of the displacment (∆r ) . Mathematically, W = F cos θ . ∆r (from Figure) r r (Applying a . b = ab cos θ ) = F .∆ r

Work is a scalar quantity. Its unit is Joule. (i) (ii) (iii) (iv) (v)

Force does not work if point of application of force does not move ( ∆ r = 0) Work done by a force is zero if displacement is perpendicular to the force (θ = 900) If the angle between the force and the displacement is acute (θ < 900), we say that work done by the force is positive. If the angle between the force and the displacement is obtuse (θ > 900), we say that work done by the force is negative. Work done depends on the frame of reference. With the change of the frame of reference inertial force does not change while displacement may change.

WORK DONE BY A VARIABLE FORCE Consider a particle being displaced along the x-axis under the action of a varying force, as shown in the figure. The particle is displaced in the direction of increasing x from x = xi to x = xf. In such a situation, we cannot use W = (F cos θ)r to calculate the work done by the force because this relationship applies only when F is constant in magnitude and direction. However, if we imagine that the particle undergoes a very small displacement ∆x, shown in the figure (1), then the x component of the force, Fx, is approximately constant over this interval, and we can express the work done by the force for this small displacement as W1 = Fx ∆x

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2 Area = ∆A = Fx ∆x

Fx

Fx

Fx xi

xf

∆x

(1)

x

xf

xi

x

(2)

Thus is just the area of the shaded rectangle in the figure (1). If we imagine that the Fx versus x curve is divided into a large number of such intervals, then the total work done for the displacement from xi to xf is approximately equal to the sum of a large number of such terms. xf

W=

∑ F ∆x x

xi

If the displacements are allowed to approach zero, then the number of terms in the sum increases without limit but the value of the sum approaches a definite value equal to the area under the curve bounded by Fx and the xaxis. xf

xf

xi

xi

lim ∑ Fx ∆x = x →0

∫ F dx x

This definite integral is numerically equal to the area under the Fx versus x curve between xi and xf. Therefore, we can express the work done by Fx for the displacement of the object from xi to xf as xf

W=

∫ F dx x

xi

when Fx = F cosθ is constant. If more than one force acts on a particle, the total work done is just the work done by the resultant force. For systems that do not act as particles, work must be found for each force separately. If we express the resultant force in the x-direction as ∑Fx, then the net work done as the particle moves from xi to xf is xf

Wnet =

∫ (∑ F )dx x

xi

Illustration: A chain of mass m = 0.80 kg and length l = 1.5 m 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 n = 1/3 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 ? Solution: Slipping occurs when the weight of hanging part is just sufficient to overcome the frictional force exerted by the table. Let µ be the coefficient of friction between chain and table. Weight of hanging part = µ (weight of horizontal part) APEX INSTITUTE - 62, Nitikhand-III, Indirapuram, Ghaziabad Ph.No.-+91-9910817866, 0120-4901457, Website: www.apexiit.co.in/

3 nmg = µ (1 – n) mg n ...(i) 1− n Let x be the length of the hanging part at some time instant. frictional force f(x) = µ (normal reaction)

µ=

µ(l − x ) mg l The work done by the frictional force if the hanging part increases to (x + dx) is : dW = – f (x) dx

=

W = ∫ dW = –

µ(l − x ) mg dx nl l

∫

l

l

µmg W=– l

 x2  x − l   2  nl 

l  2  W = – µ mg l(1 − n ) − (1 − n ) 2  

Substituting the value of µ from (i), we get : W=–

n (1 − n ) mgl = – 1.3 J. 2

WORK DONE BY A SPRING Consider the situation shown in figure. One end of a spring is attached to a fixed vertical support and the other end to a block which can move on a horizontal table. Let x = 0 denote the position of the block when the spring is in its natural length. We shall calculate the work done on the block by the spring-force as the block moves from x = 0 to x = x1. x=0

x = x1

A

A

The force on the block is k times the elongation of the spring. But the elongation changes as the block moves and so does the force. We cannot take F out of the integration ∫ F d r . We have to write the work done during a small interval in which the block moves from x to x + dx. The force in this interval is kx and the displacement is dx. The force and displacement are opposite in direction. So,

F . d r = -F dx = - kx dx

during this interval. The total work done as the block is placed from x = 0 to x1 is x

x1

1 1 2  1 2 − kxdx = − kx W= ∫  2  = − 2 kx1 0 0

If the block moves from x = x1 to x = x2, the limits of integration are x1 and x2 and the work done is

1 2

2 W =  kx1 −

1 2 kx 2  2 

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4 Note that if the block is displaced from x1 and x2 and brought back to x = x1, the work done by the springforce is zero. The work done during the return journey is negative of the work during the onward journey. The net work done by the spring-force in a round trip is zero. Illustration: A block of mass m has a velocity v0 when it just touches a spring. The block moves through a distance l before it stops after. The spring constant is k. What is the work done on it by the spring force? Solution : The net force acting on the block by the spring is equal to Fspring = – kx Work done by the spring = -∫Fspring.ds l

= - ∫ kx.dx = 0

− kl 2 2

v0

k

m v

WORK DEPENDS ON THE FRAME OF REFERENCE

Work done by a force is given as r

rr W = F.S

Since S displacement of point of application is frame dependent phenomena work is frame dependent quantity, same force may do different work in different frames Illustration : Over a horizontal plank a small block of mass m is lying at rest. Now plank is moved with constant acceleration a such that there is no relative motion between block and plank. Find the work done by friction of plank on block in first t seconds. (a) in ground frame (b) in plank frame. m a Solution: (a) In ground frame Friction force acting on the block (f) = ma Displacement in first t second S=

Work done

1 2 at 2 1 2

 

2 = F.S. = f  at 

1 2

1 2

2 2 2 = ma . at = ma t

(b)

a f

FP(=ma) f

In plank frame f = ma But displacement s = 0 (because there is no relative motion between plank and block)

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5 ∴

W = F.S. =f.s = ma . 0 =0

Note: While calculating net work done on a body in accelerated frame work done by pseudo force need to be taken into account. WORK-ENERGY THEOREM If the work done by the net force on a particle can be calculated for a given displacement, the change in the particle's speed will be easy to evaluate. As shown in the figure a particle of mass m moving to the right under the action of a constant net force F. Because the force is constant, we know Newton's second law that the particle will move with a constant acceleration a. If the particle is displaced a distance s, the net work done by the force F is Wnet = Fs = (ma)s We found that the following relationships are valid when a particle moves at constant acceleration s

m

F

vf

vi s=

v − vi 1 (vi + v f )t ; a = f 2 t

where vi is the speed at t = 0 and vf is the speed at time t. Substituting these expressions

 v f − vi  t

Wnet = m 

1  (vi + v f )t 2

1 2 1 mv f − mvi2 2 2 Wnet = Kf - Ki = ∆K Wnet =

. . . . (i)

That is, the work done by the constant force Fnet in displacing a particle equals the change in kinetic energy of the particle. Equation (i) is an important result known as the work-energy theorem. For convenience, it was derived under the assumption that the net force acting on the particle was constant. Now, we shall show that the work-energy theorem is valid even when the force is varying. If the resultant force acting on a body in the x direction is ∑Fx, then Newton's second law states that

∑Fx = ma. Thus, we express the net work done as APEX INSTITUTE - 62, Nitikhand-III, Indirapuram, Ghaziabad Ph.No.-+91-9910817866, 0120-4901457, Website: www.apexiit.co.in/

6 xf

Wnet =

xf

∫ (∑ F )dx = ∫ ma dx x

xi

xi

Because the resultant force varies with x, the acceleration and speed also depend on x. We can now use the following chain rule to evaluate Wnet. a=

dv dv dx dv = =v dt dx dt dx

Wnet =

1 2 1 mv f − mvi2 2 2

Illustration: A bullet leaving the muzzle of a rifle barrel with a velocity v penetrates a plank and loses one fifth of its velocity. It then strikes second plank, which it just penetrates through. Find the ratio of the thickness of the planks supposing average resistance to the penetration is same in both the cases. Solution: Let R = resistance force offered by the planks, t1 = thickness of first plank, t2 = thickness of second plank. For first plank : Loss in KE = work against resistance 1 1 4 mv2 – m ( v)2 = Rt1 2 2 5

⇒

1 mv2 2

 9    = Rt1  25 

.................(i)

For second plank 1 4 m ( v)2 – 0 = Rt2 2 5

⇒

 16  1 mv2   = Rt2 2  25 

.................(ii)

t1 9 dividing (I) & (II) ⇒ t = . 16 2 POTENTIAL ENERGY AND CONSERVATION OF MECHANICAL ENERGY We define the change in potential energy of a system corresponding to a conservative internal force as f

Uf - Ui = -W = -

∫ F.d r i

where W is the work done by the internal force on the system as the system passes from the initial configuration i to the final configuration f. APEX INSTITUTE - 62, Nitikhand-III, Indirapuram, Ghaziabad Ph.No.-+91-9910817866, 0120-4901457, Website: www.apexiit.co.in/

7 Note: We don't (or can't) define potential energy corresponding to a nonconservative internal force. Suppose only conservative internal forces operate between the parts of the system and the potential energy U is defined corresponding to these forces. There are either no external forces or the work done by the them is zero. We have Uf - Ui = -W = -(Kf - Ki) . . . . . (i) or Uf + Kf = Ui + Ki The sum of the kinetic energy and the potential energy is called the total mechanical energy. We see from equation (i) that the total mechanical energy of a system remains constant if the internal forces are conservative and the external forces do not work. This is called the principle of conservation of mechanical energy. The total mechanical energy K + U is not constant if nonconservative forces, such as friction, act between the parts of the system. We can't apply the principle of conservation of energy in presence of nonconservation forces. The work-energy theorem is still valid even in the presence of nonconservative forces. Note: that only a change in potential energy is defined above. We are free to choose the zero potential energy in any configuration just as we are free to choose the origin in space anywhere we like. Brain Teaser: One persons says that the potiential energy of a particular book kept in an almirah is 20 J and the other says it is 30 J. Is one of them necessarily wrong? If nonconservative internal forces operate within the system, or external forces do work on the system, the mechanical energy changes as the configuration changes. According to the work-energy theorem, the work done by all the forces equals the change in the kinetic energy. Thus, Wc + Wnc + Wext = Kf - Ki where the three terms on the left denote the work done by the conservative internal forces, nonconservative internal forces and the external forces. As Wc = -(Uf - Ui), We get Wnc + Wext = (Kf + Uf) - (Ki + Ui) = Ef - Ei . . . . . (ii) Where E = K + U is the total mechanical energy. If the internal forces are conservative but external forces also act on the system and they do work, Wnc = 0 and from (ii) . . . . .(iii) Wext = Ef - Ei The work done by the external forces equals the change in the mechanical energy of the system. Let us summarise the concepts developed so far in this chapter. (1) (2) (3)

(4) (5)

Work done on a particle is equal to the change in its kinetic energy. Work done on a system by all the (external and internal) forces is equal to the change in its kinetic energy. A force is called conservative if the work done by it during a round trip of a system is always zero. The force of gravitation, Coulomb force, force by a spring etc. are conservative. If the work done by it during a round trip is not zero, the force is nonconservative. Friction is an example of nonconservative force. The change in the potential energy of a system corresponding to conservative internal forces is equal to negative of the work done by these forces. If no external forces act (or the work done by them is zero) and the internal forces are conservative, the mechanical energy of the system remains constant. This is known as the principle of conservation of mechanical energy. APEX INSTITUTE - 62, Nitikhand-III, Indirapuram, Ghaziabad Ph.No.-+91-9910817866, 0120-4901457, Website: www.apexiit.co.in/

8 (6) (7)

If some of the internal forces are nonconservative, the mechanical energy of the system is not constant. If the internal forces are conservative, the work done by the external forces is equal to the change in mechanical energy.

Brain Teaser: When you lift a box from the floor and put it on an almirah the potential energy of the box increases, but there is no change in its kinetic energy. Is it a violation of conservation of energy? Illustration: In the figure shown stiffness 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 the maximum elongation in the spring Neglect the mass of the spring, pulley and that of the string. Solution: 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.

⇒

and

E1 = mgh1 + 0 + 0 = mgh1

m m

. . . . (i)

m

E2 = U2g + U2s + T2

1 2 = mgh2 + kx + 0 2 From conservation of energy E1 = E2

⇒

1 2 mgh1 = mgh2 + kx 2

⇒

1 2 kx = mg ( h 1 − h 2 ) = mgx 2

⇒

x = 2mg/k

1

. . . . . (ii)

h1

m

2 h2

S

Illustration: A block is placed on the top of a plane inclined at 37° with horizontal. The length of the plane is 5 m. The block slides down the plane and reaches the bottom. (a) Find the speed of the block at the bottom if the inclined plane is smooth. (b) Find the speed of the block at the bottom if the coefficient of friction is 0.25 Solution: Let h be the height of inclined plane A R ⇒ h = 5 sin 37° = 3 m (a) As the block slides down the inclined plane, it loses GPE and h vs 37º gains KE. mg cos37º mg Loss in GPE = gain in KE 37º C B mg (loss in height) = KEf – KEi APEX INSTITUTE - 62, Nitikhand-III, Indirapuram, Ghaziabad Ph.No.-+91-9910817866, 0120-4901457, Website: www.apexiit.co.in/

9 ⇒ mgh = ⇒ v=

1 mv2 – 0 2

2gh =

2 × 9 .8 × 3 = 7.67 m/s.

Note : 1. Loss in energy = initial energy - final energy 2. gain in energy = final energy - initial energy (b) As the block comes down, it loses GPE. It gains KE and does work against friction. loss in GPE = gain in KE + work done against friction A ⇒ mgh = (1/2 mv2 – 0) + (µ mg cos 37°) s R µm gc 2 o s 37 ⇒ 3mg = 1/2 mv + (0.25) × mg × 4/5 × 5 º ⇒ v=

4g = 6.26 m/s

h mg cos37º

37º

v

s

mg

37º

C B Illustration: A 1.0 kg block collides with horizontal light spring of force constant 2 N/m. The block compresses the spring 4 m from the rest position. Assuming that the coefficient of kinetic friction between the block and the horizontal surface is 0.25, what was the speed of the block at the instant of collision ?

Solution: When the block compresses the spring, let x m be the amount of compression, i.e. x = 4m. Let v = velocity of the block when it collides with the spring. Loss in KE of the block = (gain in elastic potential energy of the spring) + (work done against friction) ⇒

1 1 mv2 – 0 = kx2 + µ mg x 2 2 1 1 mv2 = (2) (4)2 + 0.25 × 1 × 9.8 × 4 2 2 2 v = 51.6 ⇒ v = 51.6 = 7.18 m/s

Illustration: A pump is required to lift 1000 kg of water per minutes from a well 20 m deep and eject it at a rate of 20 m/s. (a) How much work is done in lifting water ? (b) How much work is done in giving it a KE ? Solution: (a) Work done in lifting water = gain in PE (potential energy) work = 1000 × g × 20 = 1.96 × 105 J per minute (b) Work done (per minute) in giving it KE = 1/2 mv2 = 1/2 (1000) (20)2 = 2 × 105 J per minute

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10 Brain Teaser :

A "Violation" of the Law of the Conservation of Energy The following argument seems to prove a violation of the law of the energy conservation. Suppose that a resting hand-cart of mass m is hit by and retains a projectile of the same mass. The projectile had been flying horizontally before the collision at a velocity v in the same direction as the hand-cart. As a result of this impact the hand-cart and projectile will together set in motion at a initial velocity which can be found from the law of the conservation of momentum

v1 =

mv v = . 2m 2

Hence, the kinetic energy of the hand-cart and projectile together is 2

v 2m  2  2  = mv , W1 = 2 4 while before the collision the projectile had a kinetic energy of

mv 2 W= . 2 i.e., twice as larger. Thus, after the collision half the energy has vanished altogether. Can you say where it has gone? TYPES OF POTENTIAL ENERGY 1. Elastic We have seen that the work done by the spring force (of course conservative for an ideal spring) is -

1 2 kx when the spring is stretched or compressed by an amount x from its unstretched or natural position. 2

 1 2 kx    2

Thus,

U = -W = -  −

or

U=

1 2 kx 2

(k = spring constant)

Note that elastic potential energy is always positive. 2. Gravitational The gravitational potential energy of two particles of masses m1 and m2 separated by a distance r is given by U= −G Here,

m1m2 r

G = universal gravitation constant

N − m2 = 6.67 × 10 kg 2 -11

If a body of mass m is raised to a height 'h' from the surface of earth, the change in potential energy of the system (earth + body) comes out to be: APEX INSTITUTE - 62, Nitikhand-III, Indirapuram, Ghaziabad Ph.No.-+91-9910817866, 0120-4901457, Website: www.apexiit.co.in/

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mgh ∆U = 1 + h   R

(R = radius of earth)

∆U ≈ mgh if h m). The ring is held on a level with the peg and released : 2mMa . Show that it first comes to rest after falling a distance : 2 M − m2

2.

A small body A starts sliding from the height h down an inclined groove passing into a half - circle of radius h/2 (see figure). Assuming the friction to be negligible, find the velocity of the body at the highest point of its trajectory (after breaking off the groove).

3.

An ideal massless spring can be compressed by 1 m by a force of 100 N. This same spring is placed at the bottom of a frictionless inclined plane which makes an angle θ = 300 with the horizontal. A 10 kg mass is released from rest at the top of the incline and is brought to rest momentarily after compressing the spring 2 meters. (a) Through what distance does the mass slide before coming to rest ? (b) What is the speed of the mass just before it reaches the spring ?

4.

5.

A spring of mass m and stiffness k is fitted to a block of mass M. The system is moving with a constant velocity v on a smooth horizontal surface. If the system collides with a wall, find the maximum compression of the spring before it recoils, assuming that the total energy is conserved.

300

v

m M

A heavy particle hangs from a point O, by string of length a. It is projected horizontally with a velocity v such that v2 = (2 + 3 ) ag, show that the string becomes slack when it has described an angle cos-1(-1/ 3 ).

6.

A stone with weight ‘w’ is thrown vertically upward into the air with initial speed v0. If a constant force f due to air drag acts on the stone throughout its flight. (a) Show that the maximum height reached by the stone is h =

v 02 2g[1 + f / w ]

w −f  (b) Show that the speed of the stone upon impact with the ground is v = v 0   w + f  APEX INSTITUTE - 62, Nitikhand-III, Indirapuram, Ghaziabad Ph.No.-+91-9910817866, 0120-4901457, Website: www.apexiit.co.in/

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7.

A heavy particle hangs by a inextensible string of length ‘a’ from a fixed point and is then projected horizontally with a velocity (2gh ) . If (5/2)a > h >a, Prove that circular motion ceases when the greatest height ever reaches by the particle above the point of projection is (4a - h)(a + 2h)2/(27a2).

8.

A particle is projected, along the inside of a smooth fixed sphere, from the lowest point, with a velocity equal to the due to falling freely down the vertical diameter of the sphere. Show that the particle will leave the sphere and afterwards pass vertically over the point of projection at a distance equal to (25/32) of the diameter.

9.

Show that a particle projected with velocity

(2ag) from the lowest point of a vertical circle of radius a and moving inside it will just reach the end of the horizontal diameter; while if projected with velocity

(5ag) , it will just reach the highest point. Prove that the reaction at any point in the first case is proportional to the depth below the horizontal diameter and in the second case to the depth below the highest point. 10.

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 by v = (2pt / m)1/ 2 1/ 2

 8P   (b) The position is given as a function of time by s =   9m 

11.

t 3/ 2 .

A particle of mass m moves along a circle of radius R with a normal acceleration varying with time as a n = bt 2 , where b is a constant. Find the time dependence of the power developed by all the forces acting on the particle, and the mean value of this power developed by all the forces acting on the particle, and the mean value of this power averaged over the first t seconds after the beginning of motion.

12.

Two bars of masses m1 and m2 connected by a weightless spring of stiffness k (figure) rest on a smooth horizontal plane. x 1

2

Bar 2 is shifted a small distance x to the left and then released. Find the velocity of the centre of inertia of the system after bar 1 breaks off the wall. 13.

In a certain two-dimensional field of force the potential energy of a particle has the form U = αx 2 + β y 2 , where α and β are positive constants whose magnitudes are diferent. Find out : (a) whether this field is central: (b) what is the shape of the equipotential surfaces and also of the surfaces for which the magnitude of the vector of force F = const. APEX INSTITUTE - 62, Nitikhand-III, Indirapuram, Ghaziabad Ph.No.-+91-9910817866, 0120-4901457, Website: www.apexiit.co.in/

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14.

One end of a spring of natural length h and spring constant k is fixed at the ground and the other is fitted with a smooth ring of mass m which is allowed to slide on a horizontal rod fixed at a height h (see figure). Initially, the spring makes an angle of 37° with the vertical when the system is released from rest. Find the speed of the ring when the spring becomes vertial.

15.

A car is travelling on a level road with speed v0 at the instant when the brakes lock, so the tires slide rather than roll. (a) Use the work energy theorem to calculate the minimum stopping distance of the car in terms of v0, the acceleration of gravity g, and the coefficient of kinetic friction µk between the tires and the road. (b) The car stops in a distance of 98.3 m if v0 = 90 km/h. What is the stopping distance if v0 = 60 km/h ? Assume that µk remains the same, so that the friction force remains the same.

LEVEL - III (CHECK YOUR

SKILLS)

1.

A body with zero initial velocity slips from the top of an inclined plane forming an angle α with the horizontal. The coefficient of friction µ between the body and the plane increases with the distance l from the top according to the law µ = bl . The body stops before it reaches the end of the plane. Determine the time t from the beginning of motion of the body to the moment when its comes to rest.

2.

In the reference frame K two particles travel along the x-axis one of mass m1 with velocity v1, and the other of mass m2 with velocity v2. Find: (a) The velocity v of the reference frame k’ in which the cumulative K.E. of these particles is minimum. (b) The cumulative K.E. of these particles in the k’ frame

3.

A thin rim of mass m and radius r rolls down an inclined plane of slope α , winding thereby a thin ribbon of linear density (shown in the figure). At the initial moment, the rim is at a height h above the horizontal surface. Determine the distance s from the foot of the inclined plane at which the rim stops, assuming that the incline plane smoothly changes into the horizontal plane.

4.

r

h α

A small particle of mass m initially at A (see Figure) slides down a frictionless surface AEB. When the particle is at the point C, show that the angular velocity and the force

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44

exerted by the surface are

ω=

m A

r

2g sin α and F = 3 mg sin α r

B

α C E

5.

A chain AB of length l is loaded in a smooth horizontal tube so that its fraction of length h hangs freely and touches the surface of the table with its end B. At a certain moment, the end A of the chain is set free, with what velocity will this end of the chain slip out of the tube?

6.

Figure shows a smooth track, a part of which is a circle of a radius R. A block of mass m is pushed against a spring constant k fixed at the left end and is then released. Find the initial compression of the spring so that the block presses the track with a force mg when it reaches the point P, where the radius of the track is horizontal.

7.

Three identical spring A , B and C each of natural length l and

A

h B

spring constant K are connected to a point mass m as shown in the figure. A and B are horizontal and C is vertically fixed with rigid supports. What is the work done by the external agent in slowly lowering the mass m till it attains equilibrium when the springs A and B make an angle 2 sin–1 3/5 between them. Neglect the masses of the springs. 8.

A small sphere tied to the string of length 0.8m is describing a vertical circle so that the maximum and minimum tensions in the strings are in the ratio 3:1. The fixed end of the string is at a height of 5.8m above ground. (a) Find the velocity of the sphere at the lowest position. (b) If the string suddenly breaks at the lowest position, when and where will the sphere hit the ground? (take = 10 m/s2)

9.

A body of mass m was slowly hauled up the hill. (fig) by a force F which at each point was directed along a tangent to the trajectory. Find the work performed by this force, if the height of the hill is h, the length of its base l, and coefficient of kinetic friction k.

E

10.

A small bar resting on a smooth horizontal plane is attached by threads to a point P (fig.) and by means of a weightless pulley, to a weight B possessing the same mass as the bar itself. Besides, the bar is also attached to a point O by means of a light nondeformed spring of length l 0 = 50 cm and the stiffness K = 5 mg/l0, where m is the mass of the bar. The thread PA having

m h l

O l0

P

A

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B

45

been burned, the bar starts moving. Find its velocity at the moment when it is breaking off the plane. 11.

12.

A horizontal plane supports a plank with a bar of mass m = 1.0 kg placed on it and attached by a light elastic non-deformed cord of length l0 = 40 cm to a point O (fig.). The coefficient of friction between the bar and the plank equals k = 0.20. The plank is slowly shifted to the right until the bar starts sliding over it. It occurs at the moment when the cord deviates from the vertical by an angle θ = 300. Find the work that has been performed by that moment by the friction force acting on the bar in the reference frame fixed to the plane. A block of mass m connected with a light spring of natural length l0 and stiffness k as shown in the figure. The spring is relaxed the block is pulled by an external force slowly. Find the work done by the external agent till the block

The figure shows a ball A of mass m connected to a light spring of stiffness k. Another identical ball B is connected with the ball A by a light inextensible string as shown in the figure. Other end of the spring is fixed. Initially the spring is in relaxed position. A vertical force F acts on B such that the balls move slowly. What is the work done by the force in pulling the ball B till that ball A reaches at the top of the cylindrical surface the ball A remains in contact with the surface and coefficient of friction between the surface and the ball A is µ .

l0

m

θ

l

  mg break off the surface.  Given kl = 1 . 0   13.

O

m

m mg

F N

A R π/4

B m1

14.

A weightless horizontal rigid rod along which two ball of the same mass m can move without friction rotates at a constant angular velocity ω about a vertical axle. The ball are connected by a weightless spring of rigidity k, whose length in the undeformed state is l0. The ball which is closer to the vertical axle is connected to it by a similar spring. Determine the lengths of the springs. Under what conditions will the balls move in the circle ?

15.

Two blocks P and Q of mass 2 m and m respectively are connected by a massless string and are at rest as shown in figure all pulleys are ideal and the surface is frictionless. Find the velocity of the block P at point A and B when the system is released from rest. [at A, thread from P to pulley is vertical]

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46

LEVEL - IV (IIT-JEE PROBLEMS) (JUDGE YOURSELF AT JEE-LEVEL) O

1.

A particle is suspended vertically from a point O by an inextensible massless string of length L. A vertical line AB is is at a distance of LS from O as shown. The object is given a horizontal velocity u. At some point, its motion ceases to be circular and eventually the object passed through the line AB. At the instant of crossing AB, its velocity is horizontal. . Find u. [I.I.T. 1999]

L

A

L 8

u B

2.

A cart is moving along x direction with a velocity of 4 m/s. A person on the cart throws a stone with a velocity of 6 m/s relative to himself. In the frame of reference of the cart, the stone is thrown in yz plane making an angle of 300 with vertical z-axis. At the highest point of its trajectory, the stone hits an object of equal mass hung vertically from branch of a tree by means of a string of length L. A completely inelastic collision occurs in which the stone gets embedded in the object. Determine (a) The speed of the combined mass immediately after the collision with respect to an observer on the ground. (b) The length L of the string such that the tension in the string becomes zero when the string becomes horizontal during the subsequent motion of the combined mass. [ I.I.T. 1997 New]

3.

A spherical ball of mass m is kept at the highest point in the space between two fixed, concentric spheres A and B (see Figure). The smaller sphere A has a radius R and the space between the two Sphere B θ spheres has a width d. The ball has a diameter very slightly less d than d. All surfaces are frictionless. The ball is given a gentle O R push (towards the right in the figure). The angle made by the radius vector of the ball with the upward vertical is denoted by θ Sphere A (Shown in the figure). (a) express the total normal reaction force exerted by the spheres on the ball as a function of angle θ . (b) let NA and NB denote the magnitudes of the normal reaction forces on the ball exerted by the spheres A and B, respectively. Sketch the variations of NA and NB as functions of cos θ in the range 0 ≤ θ < π by drawing two separate graphs in your answer book, taking cos θ on the horizontal axes. [I.I.T. 2002]

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47

SOLUTION (OBJECTIVE) LEVEL - I 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15.

(c) (c) (b) (b) (a) (a) (d) (d) (c) (a) (c) (b) (a) (a) (c)

LEVEL - II 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15.

(c) (d) (c) (a) (b) (b) (d) (c) (b) (d) (c) (d) (b) (a) (a)

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48

SOLUTION (SUBJECTIVE) LEVEL - I (CBSE) 1.

(a) 882 J (c) 635 J

(b) -247 J (d) 635 J Work done by the net force on a body equals change in its K.E.

4.

12 J

5.

0.082 J in each half, – 0.163 J

6.

50 J

7.

(a) 200 m2

8.

0.125

10.

8.82 J for both cases.

11.

0.96 J

14.

(a) 800 N

(b) comparable to the roof of a large house of dimension 14 m × 14 m 9.

43.6 KW

13.

2d

(b) 12 J

LEVEL - II 2.

v=

3.

(a)

4. 11.

2 gh 3 3 4m

(b)

2 5 m/s

3M + m v 3k

P = mRat =

mRat 2

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49

12.

v c = x km 2 /( m1 + m 2 )

13.

(a)

No;

(b)

ellipses whose ratio of semiaxes is a / b = β / α ; also ellipses, but with a/b = β / α .

14.

h k 4 m

15.

V02 (a) S = 2µ k g

(b) 43.7 m

LEVEL - III 1.

t=

π gb cos α r r r m1v1 + m 2 v 2 v cm = m1 + m 2

2.

(a)

3.

h  h  mg + ρ   r −  2  sin α   S= ρr

5.

v=

2gh log e

6.

3mgR k

7.

4 l [kl − mg ] 3

(b)

K.E. =

1 m1m 2 (v1 − v 2 )2 . 2 m1 + m 2

l h

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50

8.

(a) u = 8 m/s (b) horizontal displacement = 8m sphere hits the ground 1s after breaking off the spring

9.

mg(h + kl)

10.

v = 1.7 m/s

11.

w = 0.09 Joule

12.

1 2 kl0 2

13.

 µ π 1  mgR  − +1 −  2  2 4

14.

l1 =

l0 [1 − (3mω / k) + (mω2 / k) 2 ] 2

l 0 (1 − mω 2 / k ) l2 = (1 − 3mω 2 / k ) + ( mω 2 / k 2 )

ω<

15.

3− 5 k 2 m

(10)1/2 m/s 10g (5 − 2 5 ) m/s 11

LEVEL - IV

1.

 3 3  u = gL 2 + 2  

2.

(a)

V0 = 2.5 m / s

3.

(a)

N = 3 mg cos θ – 2 mg

L = 0.32 m

(b)

NA

NB 5mg

mg

(b)

cosθ -1

2/3

+1

cosθ

-1

2/3

+1

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