A fire fighter tightens the nut on the top of a fire hydrant using a large wrench.If she applies a force of 250N perpendicular to the wrench,at a point 0.75m from the axis of rotation of the nut,how much work does she do on the nut as she rotates it through 45$^{\circ}$?
A) 190 J
B) 150 J
C) 290 J
D) 130 J

Question

Quizplus · Accepted Answer

Work done is calculated using the formula: Work = Torque × Angular Displacement. Torque = Force × Distance = 250 N × 0.75 m = 187.5 Nm. Angular Displacement in radians = 45° × (π/180) = π/4 rad. Work = 187.5 Nm × π/4 rad ≈ 147.26 J, which rounds to 150 J.

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