weight, work, kinetic energy....?

Question

A cart of mass 1.7 kg coasts 70 cm up an incline at 6 degrees with horizontal. Assume that frictional and other nongravitational forces parallel to the incline are negligible.
•What is the component of the cart's weight parallel to the incline?
•How much work does this force do as the cart rolls up the incline?
•How much work does the net force do as the cart rolls up the incline?
•Using the definition of kinetic energy determine the velocity of the cart after coasting the 70 cm, assuming its initial velocity to be zero.

Answer 1

Since the incline is frictionless, all of the work done is the gain in potential energy which is g*h

where h can be found by using the 70 cm as the hypotenuse of a triangle:

sin(6)=h/.70
solve for h

Backing up to part 1:

The parallel force,p, is:
sin(6)=p/(m*g)
p=sin(6)*1.7*9.81

This force does work for 70 cm
p*.7
(note that this is equal to m*g*h)

For the third part, we have the total work equal to the gain in potential energy.

The last part simply looks at the kinetic energy when the cart slides DOWN the slope, which is equal to the potential energy lost:

.5*m*v^2=m*g*h
v=sqrt(2*g*h)

h=.7*sin(6)

solve for v

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