Cart Pushing: what happend to s & fk?

1. Mar 13, 2006

Sakura

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:

2. Mar 13, 2006

eep

Perhaps this can be better explained, but I think of force as an instantaneous quantity. I don't think you should approach part a by thinking about work. What does the fact that the shopping cart is traveling at constant velocity tell you about the net force on the shopping cart?

As for part (c), what direction is the frictional force pointed in?

3. Mar 13, 2006

Sakura

(a) The only thing I can think of is since velocity is constant, there is no acceleration, but I dont see how acceleration effects the net force. I'm sorry, but I'm not seeing something here??

(c)In my diagram, Fk points in the left direction.

4. Mar 13, 2006

eep

(a) Think newton's laws. F = ma. So if a = 0, what does F equal? What does this tell you about Ffrction + Fpush?

(c) Why do you want to multiply by the cosine of the angle?