<p>A machine pulls a 20 kg trunk 3.0 m up a ramp that makes an angle of 30&#186;

Question

<p>A machine pulls a 20 kg trunk 3.0 m up a ramp that makes an angle of 30&#186;. The trunk maintains a constant velocity, with the machine's force on the trunk directed parallel to the ramp. The &#956;_k is .20. What are (a) the work done on the trunk by the machine's force and (b) th work done by friction on the trunk?</p>
<p>&#160;</p>
<p>I believe I should start a by finding the force of friction as well as the force of gravity acting parallel to the trunk, and set this equal and opposite the force of the machine, as there is no acceleration, thus, the net force should equal 0.</p>
<p>So, I created the equation: -F_machine = mg cos&#952; &#956; + mg sin&#952;. Then, I would multiply this answer by 3, to find the work done by the machine? Would this be a correct way to go about this? Or should I be using the Work = &#916;Energy rule? Why or why would either work/not work?</p>
<p>I'm confident that I understand part (b), which would be the force of friction, times the distance the trunk traveled, which I got to be W = -101.845.</p>
<p>Thank you for any help with this problem! Much appreciated!</p>

Explanation / Answer

since velocity is constant, net force will be zero

=> F = mgsin + mgcos where F is machines force

=> F = 20*9.81(sin30 + 0.2*cos30)

= 132.08 N

Work done by  machines force = F * S = 132.08*3 = 396.25 J

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