Work done by a force.Workisdonewhenever aforcemoves something over a distance. You can calculate the energy transferred, orwork done, by multiplying theforceby the distance moved in the direction of theforce. ... The transfer of energy is not 100% efficient and not all the energy transferred is represented by mgh

Special cases:

CaseI: When the body moves in the direction of the force. In this case, q = 0o.

from eq (i), we have

W = FScosoo

i.e. W = FS …(ii)

i.e. W = product of the force and the distance through which the body moves in the direction of the force.

This type of work is said to be positive. Therefore, work done by a force is positive is q < 90o

Examples of Positive Work:

(1) When a spring is stretched, the work done by the stretching force is positive as the force acts in the direction of displacement of the spring.

(2) When a body falls freely under gravity, the work done by gravity is positive as force (gravity) and displacement of the body are in the same direction.

(3) When a gas taken in a cylinder expands, work done by the force is positive as the displacement is in the direction of the force.

Case II:. When the force acting on a body and its displacement are in the opposite direction

In this case, q = 180o. Therefore, from eqn (i), we have

W = FS cos180o= FS(-1)

Or W = -FS …. (iii). This type of work is said to be negative. Therefore, work done by a force is negative if q > 90o.

Examples of Negative Work:

(i) The work done against friction is negative as the frictional force is always in direction opposite to displacement.

(ii) Work done in lifting a body is negative as gravitational force and displacement are in the opposite direction.

Case III:. When force and displacement are at right angle to each other.

In This case q = 90o

From eqn (i), we have

W = FS cos90o= FS . 0 = 0

Therefore, no work is done by a force which acts at right angle to the direction of displacement of the body.

Examples of Zero Work:

(i) The work done by the centripetal force acting on a body moving in a circle with uniform speed is zero as it is always at right angle to the direction of motion of the body.

W = FS cos90o

= FS (0)

= 0

(ii) A person carrying a load on his hand and walking on a horizontal road does no work as the force is acting at right angle to displacement.

Case IV: If S = 0, then from eqn. (i), we have

W = f x 0 x cosq = 0

Therefore, no work is done by because if it can not displace a body on which it acts.