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 at26.5° 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 2.30 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?
(b) What is the speed of the boat when it strikes the water again? Neglect any effects due to air resistance.

Explanation / Answer

part a:

let the surface of the lake be the referrence point for zero potential energy.

then initial total energy=initial potential energy+initial kinetic energy

=0+0.5*mass*speed^2

=0.5*285*13.5^2=25970.625 J

on the ramp, normal force=mass*g*cos(26.5)

=285*9.8*cos(26.5)=2499.5516 N

then friction force=friction coefficient*normal force=374.9327 N

length of the ramp=2.3/sin(26.5)=5.15466 m

then work done by friction =friction force*length of the ramp=1932.65 J

when the boat reaches top of the ramp, its potential energy =mass*g*height=285*9.8*2.3=6423.9 J

let speed at which boat leaves the ramp is v.

hence using energy conservation principle:

initial total energy-work done by friction=final potential energy+final kinetic energy

==>25970.625-1932.65=6423.9+0.5*285*v^2

==>v^2=123.607
==>v=11.1178 m/s

part b:

just before leaving the ramp, total energy of the boat=potential energy+kinetic energy

=6423.9+0.5*285*11.1178^2=24037.975 J

when the boat lands on the surface, its potential energy becomes 0.
hence total initial energy converts completely to potential energy

so if speed on the surface when the boat lands is v m/s,

thrn 0.5*285*v^2=24037.975

==>v=12.9879 m/s

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