Physics

Session

Work, Power and Energy - 3

Session Objectives

Session Objective

Non Uniform circular motion

Problems

Non Uniform circular motion

min tT (at top) 0 v gr

2b

max

mvT (at bottom) mg

r

2 2b

1 1mv mgr(1 cos ) mv (ii)

2 2

2 2t bAt top : v v 2gr

2mvT mgcos .............(i)

r

= 0

T

r

mg

mgcos

Non Uniform circular motion

Using (i) and (ii)

(a) Particle just completes a circle (Ttop=0)

bottom top

bottom

V 5gr V gr

T 6mg

Illustrative Problem

m

d

0.6

One end of a string of length is pivoted and a ball of mass m is attached at the other end. The ball is released from rest from a horizontal position. There is a nail at a distance d = 0.6 directly below the point of pivoting. The change in the tension in the string just after it touches the nail and just before it touches is

(a) 3 mg (b) 6 mg

(c) mg (d) 0.5 mg3

2

Solution

Mass m reaches the bottom B, where tension

2b

Amv

T mg (in a circle of radius )

B

A O

T

0.6 L

D

mg

As the loss of P.E. is mg

2 2b b

1mv mg v 2g

2

AT mg 2mg 3mg

Solution

But vb is still the same.

So new value of tension 2

bB

mvT mg

0.4

2mgmg 6 mg

0.4

TB – TA = 3 mg

At B the radius of circle

= DB

= – 0.6 = 0.4

B

A O

T

0.6 L

D

mg

Class Exercise

Class Exercise - 1

A stone is thrown from the top of a cliff whose height is H. The magnitude of initial velocity is v. Neglecting air resistance, the stone hits the ground with maximum kinetic energy if it is thrown

(a) vertically upward (b) horizontally

(c) vertically downward (d) in any direction

Solution

Final kinetic energy = Initial kinetic energy + Loss in potential energy

Initial KE in all the cases are equal.

Loss in PE = mgh is also equal in all the cases.

Final KE is then equal in all the cases and independent of direction.

Hence answer is (d)

Class Exercise - 2

The spring, which is horizontal, is supported by two identical hanging masses as shown. When both masses are released from rest at the same time, the spring is stretched by an amount x. Then the work done by the spring on each mass is

2 2

2 2

1 1(a) kx (b) kx

2 4

1 1(c) kx (d) kx

4 2

Force constant = k

m m

Solution

The gravity forces of the masses do work

on the spring, which is . So work

done by restoring force of the spring on

both masses is So work done on

each mass (they are identical) is

21kx

2

21kx

2

21kx

4

Hence answer is (c)

Class Exercise - 3

Consider the two statements:

(i) The negative of the work done by the conservative internal forces of a system is equal to the change in total energy.

(ii) Work done by external forces on a system equals the change in total energy.

(a) (i) and (ii) are correct

(b) (i) is incorrect and (ii) is correct

(c) (i) is correct and (ii) is incorrect

(d) Both (i) and (ii) are incorrect

Solution

Conservative internal forces do work in changing the configuration of the system and by definition, the negative of the work done is equal to an increase in potential energy. (i) is incorrect. External forces, by doing work on a system change its total energy. So (ii) is correct.

Hence answer is (b)

[Both explanations are of the nature of definition of the terms.]

Class Exercise - 4

A block of mass m rests on a rough inclined plane inclined at an angle to the horizontal. The plane is fixed in a lift, which moves up with a constant velocity v. The work done by the normal reaction force on the block over a time t is

(a) mg vt (b)

(c) (d)

2mg vt cos

mg vt cos 2mg vt sin

Solution

The block is in equilibrium, so total work done is zero. But individual forces (gravity force, friction, normal reaction) do work as a displacement it exists for the block.

N = mg cos

s = vt

W = Ns cos

= mgvt cos2

N

Class Exercise - 6

A projectile is fired from the top of a tower 80 m high, with a speed of 30 m/s. What is its speed when it hits the ground? (g = 10 m/s2)?

Solution

Potential energy at height h = mgh (with respect to ground)

Kinetic energy at height 2h

1h mv

2

Kinetic energy at ground level = 2g

1mv

2

Mechanical energy at h = Mechanical energy at ground level

2 2h g

1 1mv mgh mv

2 2 2 2 2

g hv v 2gh 30 2 10 80

= 900 + 1600 = 2500 gv 50 m/ s

Class Exercise - 7

A body of mass 500 g is subjected to a force of (0.8x + 20) N, where x is displacement of the particle in metres. What work is done when the particle is displaced from x = 1 m to x = 2 m?

Solution

F is along x.

dW Fdx

f

i

x

i fx

W (x to x ) F dx f

i

x

x

(0.8x 20)dx f

fi

i

x2xx

x

0.8x[20x]

2

2 2f i

if

0.8x 0.8x20x 20x

2 2

Substituting values of f ix 2m, x 1m

W = [1.6 + 40] – [0.4 + 20]

= 21.2 J

Class Exercise - 8

The bob of a simple pendulum of length 0.5 m has a speed of 2 m/s when it makes an angle of 30° with the horizontal. If acceleration due to gravity is 10 m/s2, with what speed does the bob pass the lowest position?

Solution

or loss of potential energy of mgh

2 2h

1 1mg (1 sin ) mv mv

2 2

2 2hv v 2g (1 sin30)

2 2h

1v 2 10 0.5 4 5 9 (m/ s)

2

L h

sin

H

V

Vn

Moving from H to L, the bob loses a height of h (1 sin )

v 3m/ s

Class Exercise - 9

Block A (of mass m) is pushed against a light horizontal spring of length L and spring constant K. A is resting on a smooth horizontal surface. If it is pushed so as the spring is compressed to half its length, with what speed will it leave the spring? (The spring is attached to a rigid wall)

A

Smooth surface

m

K

Solution

When the spring is restored to its original length, the block will have the maximum velocity as whole of the potential energy of the spring would have converted to kinetic energy of the block. When the edge of the block is at distance x from wall , loss in

spring energy to kinetic energy

LL x

2

2

21 L 1K K L x

2 2 2

AK

L/2 L/2x

Solution

This is equal to kinetic energy of the

Block .21mv

2

1/ 222K L

v (L x)m 4

K LWhen fully extended, x L v .

m 2

Class Exercise - 10

The mass m is pulled up from rest by a light cord, passing around a light pulley, and subjected to a constant, horizontal force F. The mass has a constant acceleration a. Using the work energy concept, relate acceleration ‘a’ to F, acceleration due to gravity (g) and mass m.

m

F

Solution

When the mass is raised by a height h Work done by the force F increases the potential energy as well as the kinetic energy of the mass m

21Fh mv mgh (i) (v is speed at height h)

2

Work done by the net force increases the kinetic energy of the mass. Net force is equal to ma.

h

F

m

Solution

21(ma)h mv

2

Using (i) Fh = mah + mgh

F = ma + mg

a = (F – mg)/m.

Thank you