MCQ Solution: Laws of Motion
A block initially at rest is allowed to slide down a frictionless ramp and attains a speed v at the bottom. To achieve a speed of 2v at the bottom, how many times as high must a new ramp be?
1) Ö2
2) 2
3) 4
4) 8
We know that the kinetic energy at the bottom of the ramp plus any friction losses sliding down the ramp sum to the potential energy at the top of the ramp. We can find the relation between the height h of the ramp and the velocity V at the bottom of the ramp. Neglecting fricion
mgh = 1/2 mV2 => V2 = 2gh
V = sqrt(2gh)
Vx = 2V = 2*Ö(2gh) = Ö (2gh*2^2*h)
So h has to be 4 times as high to double the speed.
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