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Physics
Session
Work, Power and Energy - 3
Session Objectives
Session Objective
Non Uniform circular motion
Problems
Non Uniform circular motion
T  mgcos 
mv2

.............(i)
r
Tmin (at top)  0  vt  gr
mvb2
Tm ax (at bottom) 
 mg
r
1
1
mv2  mgr(1  cos )  mvb2 (ii)
2
2
 p
r
T

At top : v2t  vb2  2gr
= 0
mgcos
mg
Non Uniform circular motion
Using (i) and (ii)
(a) Particle just completes a circle (Ttop=0)
Vbottom  5gr
Tbottom  6mg
Vtop 
gr
Illustrative Problem
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
m
(a) 3 mg
(b) 6 mg
3
(c)
mg
2
(d) 0.5 mg

0.6
d
Solution
Mass m reaches the bottom B,
where tension
mvb2
TA  mg 
(in a circle of radius )
As the loss of P.E. is mg,,,,,
1
mvb2  mg  vb2  2g
2
 TA  mg  2mg  3mg
O
A
0.6 L
D
T
B
mg
Solution
At B the radius of circle
= DB
=  – 0.6  = 0.4 
0.6 L
But vb is still the same.
D
mvb2
So new value of tension TB  mg 
0.4
2mg
 mg 
 6 mg
0.4
 TB – TA = 3 mg
O
A
T
B
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
1 2
(a) kx
2
1 2
(b) kx
4
1 2
(c)  kx
4
1 2
(d)  kx
2
Force constant = k
m
m
Solution
The gravity forces of the masses do work
1 2
on the spring, which is
. So work
2
kx
done by restoring force of the spring on
1
both masses is  kx2 So work done on
2
each mass (they are identical) is  1 kx2
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.
[Both explanations are of the nature of
definition of the terms.]
Hence answer is (b)
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
2
mg
vt
cos

(b)
(c) mg vt cos
2
mg
vt
sin

(d)
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
N
s = vt
 W = Ns cos
= mgvt cos2


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
Kinetic energy at height h 
1
mvh2
2
Potential energy at height h = mgh
(with respect to ground)
1
2
mv
Kinetic energy at ground level =
g
2
Mechanical energy at h = Mechanical energy at ground
level
1 2
1
2
2
2
 mvh  mgh  mv2

v

v

2gh

30
 2  10  80
g
g
h
2
2
= 900 + 1600 = 2500
 v g  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
 W (xi to x f ) 
x
xf
 F dx
xi
 0.8x 2  f
x

  [20x]x f
i
 2  x
i

xf
 (0.8x  20)dx
xi
 0.8x 2
  0.8x 2
f
i  20x

 20x f   
i
 2
  2


Substituting values of xf  2m, xi  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
Moving from H to L, the bob loses
a height of h  (1  sin )
or loss of potential energy of mgh
1 2 1 2
 mg (1  sin )  mv  mvh
2
2
 v2  vh2  2g (1  sin30)
 sin 
 1
2
 vh  2  10  0.5    4  5  9 (m / s)2
 2
v  3m/ s
H
Vn

L
V
h
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)
K
A
Smooth surface
m
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 L  x  L  , loss in


2 
spring energy to kinetic energy
K
2
1 L 
1
2
 K    K L  x 
2 2
2
A
L/2
x
L/2
Solution
This is equal to kinetic energy of the
1
2
Block  mv .
2
K  L2
1/ 2

v
  (L  x) 
m  4

2
When fully extended, x  L  v 
K L
.
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.
F
m
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
F
1 2
 Fh  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
m
Solution
1
 (ma)h  mv 2
2
Using (i) Fh = mah + mgh
F = ma + mg
a = (F – mg)/m.
Thank you