# Problem 7-35 Work-energy theorem - Part 2 - E

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**

Correct!

\(\sum F_d = ½ \;m(v - v_0)^2\) is incorrect. The correct ones are:

(A) \(\text{Work done = Change in }E_K\)

(B) \(\sum F_d = \Delta(½ \;mv^2)\)

(C) \(\sum F_d = \Delta E_K\)

(D) \(\sum F_d = ½\;m(v^2 - v_0{^2})\)

The work done \(\sum Fd\) is given by:

(A) \((95.6\; N - 75.5 \;N)(0.75\; m)\)

(B) \((95.6 \cos 35\; N - 75.5\; N)(0.75\; m)\)

(C) \((95.6\; N + 75.5\; N)(0.75\; m)\)

(D) \((95.6 \cos 35\; N + 75.5\; N)(0.75 \;m)\)