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## 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! :)

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