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