(adsbygoogle = window.adsbygoogle || []).push({}); Cart Pushing: what happend to s & fk??

A person pushes a 16.0 kg shopping cart at a constant velocity for a distance of 22.0 m. She pushes in a direction 29 degrees below the horizontal. A 48.0 N frictional force opposes the motion of the cart. (a) What is the magnitude of the force the shoper exerts? Work done by (b) pushing force, (c) friction force and (d) gravitational force?

(a) I used the formula W=(FcosΘ)s = F(cos29)-Fk = 54.9 N

My reasoning was to take the sum of the two forces (the force minus the frictional force), multiply by cosΘ and solve for F. But where did s (distance) go? I don't understand why it's not used?

(c) Work done by frictional force = -1060 J

I thought the force in this problem was 48 N. And so, I fugured W = 48(cos29)(22) Why is this not correct?

:uhh:

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# Cart Pushing: what happend to s & fk?

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