Okey, now the answers are:5 A cyclist and her bicycle have a combined mass of 70 kg. The cyclist ascends a straight hill AB of constant slope, starting from rest at A and reaching a speed of 4ms−1 at B. The level of B is 6m above

the level of A. For the cyclist’s motion from A to B, find

(i) the increase in kinetic energy

(ii) the increase in gravitational potential energy

During the ascent the resistance to motion is constant and has magnitude 60 N. The work done by the

cyclist in moving from A to B is 8000 J.

(iii) Calculate the distance AB.

560J

4120J

And 55.4 meters.

Looking at the markscheme i know exactly how they got it but not WHY and also why my methods didnt work.

So for the first one i did:

Increase in kinetic energy =

$\displaystyle

0.5mv^2 = 0.5u^2 + mgh

$

Which is 4120J

Now this is WRONG and i don't understand why. I understand that it means im working out the potential energy but i thought due to conservation the gain in potential energy would be equilivent to the gain in kinetic energy?

Well part 2 i worked out correctly although it gave me the same as my part 1 lol.

And part three i have more confusion .

Work done i thought in this case = increase in P.E + 60D

However it is meant to be:

W.D = Increase in K.E + Potential energy + 60D

EDIT: I think i understand this part now actually. Your speed increases therefore K.E increases. Your P.E increases as you go up and 60D against resistance...sorry!

And also, if you're going up a hill...surely the angle of the hill should be taken into consideration?

Where D is the distance.