college physics got me down :(?

Question

OK if you can help me up with this i would be very grateful. A person pushes a cart at a constant velocity a distance of 22.0m. She pushes in a direction 29.0 degrees below the horizontal. A 48N frictional force opposes the direction of the cart. what is the magnitude of the force that the shopper exerts? determine the work done by the (a) pushing force, the (b)frictional force and (c) the gravitational force.

Answer 1

The work done by the frictional force is simple :F x d

W(friction) = 48 x 22 = 1056N

The work done by gravity will be dependent on the mass of the cart. We have not been given this mass, nor do we know
the frictional coefficient of the surface of the ramp, which we could use to calculate this mass. Therefore, it will not be possible to provide an exact numeric solution to parts (a) and (c).

Define 'k' to be the dynamic frictional coefficient.

We can calculate the mass M of the cart in terms of 'k'

Frictional Force = (force perpendicular to ramp) x (coefficient of friction)

48 = Mg.cos(29).k

M = 48/( gk.cos(29) )

The gravitational force can only do work by changing the height of the cart in a gravity field.

Therefore, the work done by gravity can be calculated:

W(gravity) = Mg.h = M.g22sin(29)
= 48.22.gsin(29)/gk.cos(29) Joules

W(gravity) = 1056.tan(29)/k Joules

Finally, the work done by the pushing force is the difference between the other two sources of work
W(pushing) = W(friction) - W(gravity)
= 1056 - 1056.tan29/k

W(pushing) = 1056( 1 - tan29/k)

An interesting observation is that a k value of tan(29) (about 0.55) then no pushing force will be required; the force due to gravity will exactly equal the opposing force due to friction. Therefore no work will be done by the pushing force.

For a k value below 0.55, gravity will actually pull the cart down the ramp, and a force will have to be supplied in the opposite direction (up the ramp) to keep the speed constant.


Example
------------

Set: k = 0.6
Calculate: M = 48/gkcos(29)
M =9.32kg

W(gravity) = Mgh = 9.32x9.81x22.sin(29)
= 975J

W(friction) = 22x48 = 1056 Joules

W(pushing) = 1056 - 975 = 81 Joules

Answer 2

I'm pretty sure you need more information.

Suppose the cart has zero mass. In that case, the pushing force equals the friction, and they each do 22*48 Nm of work, and gravity does no work.

Suppose the pushing force is zero. In that case, the trigonometry reveals that the object must be 10 kg to exactly to counteract the friction. Gravity and friction again each do 22*48 Nm of work, with the push doing nothing.

We do know that the pushing force could not be greater than 48 N or the velocity increases rather than being constant. Similarly, the cart can't be greater than 10 kg, or the friction isn't enough to hold it back.

We need to know the mass, or the ratio of pushing force to friction force. The time taken is irrelevant.

Answer 3

Positive and negative signs of work indicate whether the object exerting the force is transferring energy to some other object, or receiving it. A baseball pitcher, for example, does positive work on the ball, but the catcher does negative work on it.
1.) 48/cos29 = 54.8N
a.) 48(22.0) = 176Nm
b.) 48(-22.0) = -176Nm
c.)The gravitational force does no work because it is not in the direction of motion.