How do I calculate height if the mass is 10.0kg, the work is 4.00x10^2Joules up an incline of unknown angle?

Question

Let’s push a crate of mass 10.0 kg slowly (at constant speed) up a slope with a coefficient of friction µ. The work we
need to achieve this is 4.00 × 10^2 J. By pulling the crate from this point back down the slope to the original position
we do work of 2.50 × 10^2 J.
The force of both pushing and pulling is parallel to the direction of the slope.
(a) Compare the pushing and pulling forces required.
(b) Consider the work done by the person in moving the crate, and the changes in energy of the crate to calculate:
(i) the amount of energy dissipated by friction as the crate slides either up or down the slope;
(ii) the height difference the crate moves through when sliding along the slope.

Answer 1

Let's set some unknowns that are constants
First, the height of the object above the horizontal plane wher the starting point is 0. Let's call the height h.

The length of the slope that we move the crate on, let's call this L.

Since the velocity is stated to be constant, thee is no acceleration of the object, so the two forces are gravity:
m*g, in the vertical direction downward,
and friction
m*g*µ
which always opposes the motion of the crate parallel to the surface of the slope.

Using conservation of energy, up the slope, the work done is equal to the gain in potential energy PE, less the work done by friction, FE

PE=400-FE
On the downward motion, the work done is equal to the frictional energy less the potential energy lost
250=FE-PE

Now I have two equations and two unknowns.

substituting PE from the first into the second,
250=FE-400+FE
650/2=FE
FE=325 J
This is the same in either direction
The upward force will be
m*g*(sin(th) +µ)
and the downward force will be

m*g*(µ-sin(th))

We must know th in order to compute the forces
to compute th we either have to know µ or L
as sin(th)=.75/L

PE=400-325
PE=75 J

recall that PE=m*g*h
using g=10
h=75/(10*10)
h=.75 m

j

Answer 2

It can't be done because your total work is the sum of the vertical vector and the horizontal vector. The distance will change because there are three separate factors to be calculated.

The first is resistance due to friction over the length of the incline.
The second is overcoming the force of gravity (vertical.)
The third is the work necessary to move the mass a horizontal distance.

Since we already know the end value, if one of these variables change, at least one of the other variables must also change in order for the end result to remain constant.
.

Answer 3

F1*s = 400 N-m
F2*s = 250 N-m
F1/F2 = 8/5

W1 = Wf + (10)(9.8)h = 400
W2 = Wf - (10)(9.8)h = 250
2Wf = 650

Wf = 325 N-m
h = (100 - 325)/(10*9.8) = 0.7653 m