A 285-kg stunt boat is driven on the surface of a lake at a constant speed of 13

Question

A 285-kg stunt boat is driven on the surface of a lake at a constant speed of 13.5 m/s toward a ramp, which is angled at 30.0° above the horizontal. The coefficient of friction between the boat bottom and the ramp's surface is 0.150, and the raised end of the ramp is 1.70 m above the water surface.

(a) Assuming the engines are cut off when the boat hits the ramp, what is the speed of the boat as it leaves the ramp?
m/s

(b) What is the speed of the boat when it strikes the water again? Neglect any effects due to air resistance.
m/s

Explanation / Answer

(a) the work done by the friction is mu*N*l, with N given by the component of the boat's weight perpendicular to the ramp (mgcosa, a being the ramp's angle); and l being the ramp's length.

l = h/sina h - ramp's height (1.70m)

so we have:

friction work = -mu*mgcos(a)*h/sin(a) = -mu*mgh/tan(a)

from conservation of energy + work done, we get:

K0 = 1/2m(v0)^2 - mu*mgh/tan(a) = 1/2m(vf)^2 + mgh (v0 is the initial velocity, 13.5 m/s)

from here you can find the final velocity vf.

(b) when the boat strikes the water again, it's potential energy mgh is converted back into kinetic energy, so denoting the velocity at water-strike v_w, we have:

1/2m(v_w)^2 = 1/2m(vf)^2 + mgh = 1/2m(v0)^2 - mu*mgh/tan(a)

enter the numbers and find the answer

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