almost impossible question (i think) due tommorow?

Question

A child riding a bike has a total mass of 40.0 kg the child approaches the top of a hill that is 10 m high and 100m long at 5.0m/s. If the force of friction between the bicycle and the hill is 20N, what is the childs velocity at the bottom of the hill? (Disregard air resistance g= 9.81 m/s^2)

A. 5.0m/s
B. 10.0 m/s
C. 11m/s
D. Child stops before reaching the bottom.





i have already posted this question and thanks to the guy who answered but i need some more answers so i can decided what i'm going to do.

why would you use 10 for g?

Answer 1

The velocity can be computed using conservation of energy

Kinetic energy at the bottom
1/2 * m* v^2
=20*v^2
will equal the potential energy lost:

m*g*h
=40*9.81*10

minus the work done by friction:

F*d
=20*100

The answer is B

BTW: The exact number is 9.81 m/s
If you use 10 m/s for g then you get 10 exactly

j

Answer 2

Oh well, here's another answer then.
5+sqrt(98.1) m/sec assuming that you can neglect friction between his/her brakes and the wheel ( what kid would put their brakes on? it's just mad) and in the hubs and that the wheels are fairly light so you don't need to account for their angular momentum then it's just the extra speed picked up by dropping 10 metres.

This doesn't seem to be in the list though so I'd go for D. 20N is not enough to stop the wheels slipping out sideways and the child will skid and crash long before it reaches the bottom.

Answer 3

Energy conservation: E=E1+E2+E3; E=m*u^2/2 is his initial kinetic energy at the bottom, u is initial speed to be found; E1=mgh is energy lost due to gravity, h=10m; E2=f*L is energy lost due to friction, f=20N, L=100m; E3=m*v^2/2 is energy still on the top, v=5m/s, m=40kg;
Thus m*u^2/2 = mgh + f*L +m*v^2/2 or u=sqrt(2gh + 2f*L/m +v^2) = sqrt(2*9.81*10 +2*20*100/40 + 5*5) = 17.92 m/s – is correct answer!

Answer 4

c

Answer 5

B