The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative: (a) work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket. (b) work done by gravitational force in the above case, (c) work done by friction on a body sliding down an inclined plane, (d) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity, (e) work done by the resistive force of air on a vibrating pendulum in bringing it to rest.
a)
The formula to calculate the work done is,
W = F s cos θ …… (1)
Here, the amount of force applied on the body is F , the displacement of the body is s and the angle between the force applied and the displacement is θ .
As the man lifts the bucket out of a well by means of rope that is tied to the bucket, the angle between the direction of force and the direction of displacement is zero, that is θ = 0 ° .
Substitute the values in the equation (1).
W = F s cos 0 ° = F s
Thus, the work done in this case is positive.
b)
As the man lifts the bucket out of a well by means of rope that is tied to the bucket, the angle between the direction of gravitational force and the direction of displacement is opposite, that is θ = 180 ° .
Substitute the values in the equation (1).
W = F s cos 180 ° = − F s
Thus, the work done in this case is negative.
c)
The friction is applied on the body sliding down an inclined plane. As the body slides down, it accelerates downward and the force of friction is applied opposite to the force. So, the angle between the direction of frictional force and the direction of displacement is opposite, that is θ = 180 ° .
Substitute the values in the equation (1).
W = F s cos 180 ° = − F s
Thus, the work done in this case is negative.
d)
The force is applied on a body that moves on a rough horizontal plane with uniform velocity. So, the angle between the applied force and the direction of displacement is the same, that is θ = 0 ° .
Substitute the values in the equation (1).
W = F s cos 0 ° = F s
Thus, the work done in this case is positive.
e)
The resistive force acts on a vibrating pendulum in bringing it to rest. As the pendulum vibrates the resistive force acts in the direction opposite to the motion of the pendulum. So, the angle between the direction of resistive force and the direction of displacement is opposite, that is θ = 180 ° .
Substitute the values in the equation (1).
W = F s cos 180 ° = − F s
Thus, the work done in this case is negative.
a)
The formula to calculate the work done is,
Here, the amount of force applied on the body is
As the man lifts the bucket out of a well by means of rope that is tied to the bucket, the angle between the direction of force and the direction of displacement is zero, that is
Substitute the values in the equation (1).
Thus, the work done in this case is positive.
b)
As the man lifts the bucket out of a well by means of rope that is tied to the bucket, the angle between the direction of gravitational force and the direction of displacement is opposite, that is
Substitute the values in the equation (1).
Thus, the work done in this case is negative.
c)
The friction is applied on the body sliding down an inclined plane. As the body slides down, it accelerates downward and the force of friction is applied opposite to the force. So, the angle between the direction of frictional force and the direction of displacement is opposite, that is
Substitute the values in the equation (1).
Thus, the work done in this case is negative.
d)
The force is applied on a body that moves on a rough horizontal plane with uniform velocity. So, the angle between the applied force and the direction of displacement is the same, that is
Substitute the values in the equation (1).
Thus, the work done in this case is positive.
e)
The resistive force acts on a vibrating pendulum in bringing it to rest. As the pendulum vibrates the resistive force acts in the direction opposite to the motion of the pendulum. So, the angle between the direction of resistive force and the direction of displacement is opposite, that is
Substitute the values in the equation (1).
Thus, the work done in this case is negative.
