Question

A fire fighter tightens the nut on the top of a fire hydrant using a large wrench. If she applies a force of 250 N perpendicular to the wrench, at a point 0.75 m from the axis of rotation of the nut, how much work does she do on the nut as she rotates it through 45o?

Answer 1

The torque acting on the nut is given as,

`\tau=Fr`

The work done on the nut can be given as,

`W=\tau\theta`

Substitute the known expression,

`W=Fr\theta`

Substitute the known values,

`\begin{gathered} W=(250N)(0.75m)(45^(\circ))(\frac{3.14\text{ rad}}{180^(\circ)}) \\ =147.2\text{ J} \end{gathered}`

Thus, the work done by person to rotate the nut is 147.2 J.