Question

please help with cart thing

A 45 kg cart is pushed up a ramp a length of 5.8 m from rest, attaining a speed of 2.6 m/s
at the top of the ramp, which is 1.7 m high. The coefficient of friction between the cart
and the ramp is 0.13.

a) Determine the work done against:
1) gravity 2) inertia 3) friction.
b) What force was used to push the cart?
c) What power was used to move the cart?

Answer 1

s = slope angle
sin s = 1.7/5.8
so s = 17 deg
so cos s = .956

1) m g h = 45 * 9.81 * 1.7

2) (1/2) m v^2 = (1/2) (45)(2.6)^2

3) normal force = m g cos T = 45*9.81*.956 = 422 N
friction force = .13 * 422 = 54.9 N
work against friction = 54.9*5.8

b) Total work done = increase in potential energy (1) + increase in kinetic energy (2) + work against friction (3)
so
F = that total work done / 5.8

c) power = total work done/ time up ramp
get time from average speed which is 2./2 = 1.3 m/s
t = 5.8 / 1.3
so
power = that total work done above /t

Answer 2

thanks for your help :)