Problem 7-35 Work-energy theorem - Part 4 - A

A girl pulls a box of mass \(20.8\; kg\) across the floor. She is exerting a force on the box of \(95.6\; N\), inclined at \(35.0^\circ\) above the horizontal. The kinetic friction force on the box has a magnitude of \(75.5\; N\). Use the work-energy theorem to determine the speed of the box after being dragged \(0.750\; m\), assuming it starts from rest. [Ans. \(0.45\; m/s\)]

Accumulated Solution

FBD with all forces, directions and angles indicated.

\(\sum F_d = ½\; m(v^2 - v_0{^2}) \\ \sum F_d = (95.6 \cos 35 \;N - 75.5 \;N)(0.75 \;m) = 2.108 \;J\)

Correct!

The change in EK is:

\(½ (20.8 \;kg)v^2 - ½ (20.8)0^2 = ½ (20.8 \;kg)v^2\)

and by the Work-energy theorem this is \(= 2.108\)

\(v^2 = 0.2027 \\ v = 0.45 \;m/s\)

 

You have completed this problem.