Chapter 2 WORK, ENERGY AND
POWER
1.
(i)Define work
(ii)What are the factors on which the amount of work done by a force on a body depends?
(iii)Name and define the C.G.S and S.I units of work and derive a relationship between them.
ANSWERS:
(i)Work is said to be done only when the force applied on a body makes the body move (i.e., there is
a displacement of the body) and the amount of work done by a force is equal to the product of the
force and the displacement of the point of application of the force in the direction of force.
W=F × S
(ii)The factors on which the amount of work done by a force depends are:
(a)Force
(b)Displacement of the point of application of force in the direction of force
(iii)The C.G.S unit of work is erg. 1 erg of work is said to be done when a force of 1 dyne displaces a
body through a distance of 1 cm in its own direction.
The S.I unit of work is joule. 1 joule of work is said to be done when a force of 1 newton displaces a
body through a distance of 1 meter in its own direction.
1 joule=1N × 1m
But, 1N=10 5 dyne and 1m=102 cm
! joule=105 dyne × 102 cm
107dyne × cm
= 107 ergs
Thus, 1 joule=107 erg
2.Work done by a force when the displacement is not along the force by taking the component of
displacement along the force.
C
F
, F
In ∆ABC
AB/AC=cos Ө
AB/S=cos Ө
AB= S cos Ө
Work depends on:
(i)the magnitude of force
(ii)the magnitude of displacement
(iii)the angle between the force and displacement
3.
Special cases:
(i)case I- When displacement is in the direction of force [i.e., Ө=0 °]
Cos 0°=1; W=F × S cos Ө =F × S × 1
=F × S
W=F × S
Here, work done is maximum and positive.
E.g., - If a body falls freely under gravity, it does positive work with respect to the force of gravity.
(ii)case II – When displacement is normal or perpendicular to direction of force (i.e., Ө=90 °)
Cos 90°=0
W= F ×S cos 90°
F×S×0
=0
Here, work done is zero.
E.g., -A man carrying a load and walking on a horizontal road does no work or zero work with respect
to force of gravity.
A planet going around the sun does no work or zero work. The displacement in any instant is
perpendicular to the centripetal force required for motion.
(iii)case III-When displacement is opposite to direction of force [i.e., Ө=180 °]
W=F × S × cos Ө
=F × S × (-1)
=-F × S
Here, work done is negative.
POWER
1.
(i)Define work
(ii)What are the factors on which the amount of work done by a force on a body depends?
(iii)Name and define the C.G.S and S.I units of work and derive a relationship between them.
ANSWERS:
(i)Work is said to be done only when the force applied on a body makes the body move (i.e., there is
a displacement of the body) and the amount of work done by a force is equal to the product of the
force and the displacement of the point of application of the force in the direction of force.
W=F × S
(ii)The factors on which the amount of work done by a force depends are:
(a)Force
(b)Displacement of the point of application of force in the direction of force
(iii)The C.G.S unit of work is erg. 1 erg of work is said to be done when a force of 1 dyne displaces a
body through a distance of 1 cm in its own direction.
The S.I unit of work is joule. 1 joule of work is said to be done when a force of 1 newton displaces a
body through a distance of 1 meter in its own direction.
1 joule=1N × 1m
But, 1N=10 5 dyne and 1m=102 cm
! joule=105 dyne × 102 cm
107dyne × cm
= 107 ergs
Thus, 1 joule=107 erg
2.Work done by a force when the displacement is not along the force by taking the component of
displacement along the force.
C
F
, F
In ∆ABC
AB/AC=cos Ө
AB/S=cos Ө
AB= S cos Ө
Work depends on:
(i)the magnitude of force
(ii)the magnitude of displacement
(iii)the angle between the force and displacement
3.
Special cases:
(i)case I- When displacement is in the direction of force [i.e., Ө=0 °]
Cos 0°=1; W=F × S cos Ө =F × S × 1
=F × S
W=F × S
Here, work done is maximum and positive.
E.g., - If a body falls freely under gravity, it does positive work with respect to the force of gravity.
(ii)case II – When displacement is normal or perpendicular to direction of force (i.e., Ө=90 °)
Cos 90°=0
W= F ×S cos 90°
F×S×0
=0
Here, work done is zero.
E.g., -A man carrying a load and walking on a horizontal road does no work or zero work with respect
to force of gravity.
A planet going around the sun does no work or zero work. The displacement in any instant is
perpendicular to the centripetal force required for motion.
(iii)case III-When displacement is opposite to direction of force [i.e., Ө=180 °]
W=F × S × cos Ө
=F × S × (-1)
=-F × S
Here, work done is negative.
