Work and kinetic energy on cart going down ramp.?

Question

Not looking for answers just looking for a walk through on any of these questions.

A cart of mass 1.5 kg is given an initial velocity of 2.1 m/s up a uniform incline at an angle of 3.3 degrees with horizontal. Assuming that the only forces acting on the cart are gravitational and normal forces, determine the following:

*How much work is done by gravity on the cart, and how much work does the cart do against gravity, as the cart travels 3.25 meters along the incline?
*By how much does its kinetic energy change during this displacement?
*Using energy considerations determine how fast the cart is moving at the end of this displacement.
*How are work done by the cart and its kinetic energy change related?
*How would the quantities calculated in this problem differ if the cart had ended up 3.25 meters down the incline?

Answer 1

the cart gains potential energy of
sin(3.3)*m*g*distance traveled on the incline
That energy is the work done by and against gravity. It is also the loss in kinetic energy

Use
.5*m*(vi^2-vf^2)=
m*g*sin(3.3)*d
and solve for vf

going down the incline the cart would lose potential energy and gain kinetic, so
.5*m*(vf^2-vi^2)=
m*g*sin(3.3)*d

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