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 45°?
A) 190 J
B) 150 J
C) 290 J
D) 130 J

Question

Quizplus · Accepted Answer

The work done on the nut can be calculated using the formula $W = \tau \theta$, where $W$ is the work, $\tau$ is the torque, and $\theta$ is the angle of rotation in radians. The torque $\tau$ is given by $\tau = F \times r$, where $F = 250 \, \text{N}$ is the force applied and $r = 0.75 \, \text{m}$ is the distance from the axis of rotation. The angle $\theta$ must be converted to radians: $\theta = 45^\circ = \frac{45}{180} \pi = \frac{\pi}{4} \, \text{radians}$. Thus, $\tau = 250 \, \text{N} \times 0.75 \, \text{m} = 187.5 \, \text{Nm}$, and $W = 187.5 \, \text{Nm} \times \frac{\pi}{4} \approx 147.65 \, \text{J}$, which rounds to 150 J.

A Fire Fighter Tightens the Nut on the Top of a Fire