Work  (Page 2/3)

• The force applied is $F$ =15 N.
• The distance moved is $s$ =20 m.
• The applied force and distance moved are in the same direction. Therefore, $F_{\parallel }$ =15 N.

These quantities are all in the correct units, so no unit conversions are required.

• We are asked to find the work done on the box. We know from the definition that work done is $W=F_{\parallel }s$

Remember that the answer must be positive as the applied force and the motion are in the same direction (forwards). In this case, you (the pusher) lose energy, while the box gains energy.

• The force applied is $F$ =40 N
• The distance moved is $s$ =30 cm. This is expressed in the wrong units so we must convert to the proper S.I. units (meters):
  $$\begin{array}{ccc}\hfill 100\mathrm{cm}& =& 1\mathrm{m}\hfill \\ \hfill 1\mathrm{cm}& =& \frac{1}{100}\mathrm{m}\hfill \\ \hfill \therefore 30\times 1\mathrm{cm}& =& 30\times \frac{1}{100}\mathrm{m}\hfill \\ & =& \frac{30}{100}\mathrm{m}\hfill \\ & =& 0,3\mathrm{m}\hfill \end{array}$$
• The applied force and distance moved are in opposite directions. Therefore, if we take $s$ =0.3 m, then $F_{\parallel }$ =-40 N.

• We are asked to find the work done on the car by you. We know that work done is $W=F_{\parallel }s$

Again we have the applied force and the distance moved so we can proceed with calculating the work done:

Note that the answer must be negative as the applied force and the motion are in opposite directions. In this case the car does work on the person trying to push.