Topic on Energy, work and power

Jun 2010
1
0
Hello there!

Question:

i) A car C of mass 1200kg climbs a hill of length 500m at a constant speed. The hill is inclined at 6 degrees to the horizontal. The driving force is exerted by C's engine has magnitude 1800N. Find the work done against resistance to the motion of C, as it climbs from the bottom of the hill to the top.
ii) Another car D, also of mass 1200kg, climbs the same hill with increasing speed. The speed at the bottom is 8ms and the speed at the top is 20ms. Assuming the resistance to motion of D is constant and has a magnitude 700N, find the work done by D's engine as D climbs from the bottom of the hill to the top.
iii) The driving force exerted by D's engine is 4 times as great when D is at the top of the hill as it is when D is at the bottom. Find the ratio of the power developed by D's engine at the top of the hill to the power developed at the bottom.

My attempt:
Part i) and ii) I get it but for part Iii) I've no idea how to get the answer. I tried:

Power at top : Power at bottom
Fv : 4 Fv
1180000 x 20 : 4(1180000)(8)
5 : 8

Thanks in advance!! :)
 
Oct 2010
63
12
Sri Lanka
i) Work done by the engine = work done against the friction +work done against the gravity

Fs = mgh +W

ii) w.d.b. engine = kinetic energy at the end - kinetic energy at the begining + work done against the gravity+ work done by the friction
E = 1/2m(v^2 - u^2) + F'. s + mgh

iii) you've got the final answer