Alright, here's the question:

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 29.0-degrees below the horizontal. A 48.0-N frictional force opposes the motion of the cart. (a) What is the magnitude of the force that the shopper exerts? Determine the work done by (b) the pushing force, (c) the frictional force, and (d) the gravitational force.

I know this is a relatively simple problem, but I missed most of class when we were discussing this stuff. So if someone could start me off on the right foot, I'm sure I'll be able to figure it out.

What I tried so far was drawing a free-body diagram then making net force equations for the X and Y directions--but I haven't had any luck doing that. I must be missing something? Either that or I'm going about this completely the wrong way.

Thanks!

Constant velocity suggests that there's no net force. Find the magnitude of the applied force such that this condition is met. You'll have to separate the forces into x and y components, but just remember that the normal force will adjust itself to cancel any force going into the ground. This means you'll really only need to consider the x direction.

--J

A quick reply is what your looking for so verify the information I give you as I may error in answering quickly..... :surprised

Draw out your freebody diagram again and get all the values figured out for your Fx, Fy, Friction, N, mg, ect.

Looks like in this case you will have the following....

N = mg + Fy
Fx = Fcos(theta)
Fy = Fsin(theta)

W = Fd if parallel so we know that W = Fx * d or W=Fxd = Fdcos(theta)

Hopefully that will get you started. I know it is just formulas you probably already know.... but....

Thanks for the help!

b)w(friction) should be force of friction*displacement..since the angle is zero
c) work done by gravity is zero because there is no up or down movement of the cart
a) F(applied)=mgcos(theta)
d) w=F(applied)*d

I think. LoL