# Work of a car's engine as it moves up a slope

## Homework Statement

A car of mass 900 kg accelerates up a slope. The velocity of the car at the bottom of the slope is 5 m/s. By the time the car reaches the top of the slope, its velocity is 15 m/s. The slope is 8m high from the ground and has a length (radius - angled length - not base) of 500m. The average frictional force experienced by the car between A and B is 50N.

Calculate: The total work done by the car's engine between the bottom and top of the slope.

## Homework Equations

F = ma
m = mass in kilograms
a = acceleration in m/s
F = force in newtons
W = F x s
s = distance
W = work

## The Attempt at a Solution

I am quite at a loss with this question but this is my attempt.
Fbottom = m x a = 900 x 5 = 4500N

Work (bottom) = f x s
= 4500 x 500 (slope is 500m)
= 2250000J

Ftop = F x ma
= 900 x 15
= 13500N
Work (top) = f x s
= 135000 x 500
= 6750000J

W (top) - W (bottom) = 450000J

I am sure that my answer is incorrect but I am unsure as to how to solve it and the memo is missing the answer, any help would be appreciated. Sorry for the n00b question! :)

Last edited:

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CAF123
Gold Member
its acceleration is 15 m/s
There is a typo here somewhere, you mean its velocity is 15m/s?

CAF123
Gold Member
It is difficult to follow your solution because you have inputted velocities in place of accelerations.
I think this is meant to be solved using the work-energy theorem, but also incorporating the effects of an external force, ie friction, so; $$W = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 + F_f d = ΔK.E + F_f d,$$ where $F_f$ is the force of friction.

CWatters
Homework Helper
Gold Member
Don't forget the gain in PE as well as KE?

CAF123
Gold Member
In taking into account the change in potential energy, i believe we add on, to the eqn above;
$$W = \vec{F}.\vec{d} = mgdcos(90 +θ) = -mgdsinθ$$

Apologies for forgetting that earlier, I must have done the problem ignoring the incline!