Question

A car of mass 0.9 tonnes is driven 200 m up a slope inclined at 5° to the horizontal. There is a resistance force of 100 N.
(i) Find the work done by the car against gravity.
(ii) Find the work done against the resistance force.
(iii) If the car slows down from 12 m s− 1 to 8 m s− 1, what is the total work done by the engine?

Here's my work:
i) Work done = force x distance
= 9000 x 20sin5
= 157000 J (3 s.f.)
ii) Work done = force x distance
= 100 x 200
= 20,000 J
iii) The loss in kinetic energy = 1/2mu^2-1/2mv^2
(1/2 x 900 x 8^2) - (1/2 x 900 x 12^2)
= -36000 J

And that's all I got, can anyone help me with the last part of this question, thank you so much!

Answer 1

(i) .9 * 1000 = 900 not 9000
are you assuming g = 10 ?
If so then indeed weight = 9000 N
it is 200 sin 5
I get 157,000 J so you just have typo

(ii) agree 20,000

for (iii)
total work done by engine = work against gravity + work against friction - loss in Ke
= 157,000 + 20,000 - 36,000 Joules