Physics problem help ---- Finding speed based upon friction, angle and height?

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 28.8° 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 1.90 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

Answer 1

(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

Why assume the engines are cut?  The boat's in air on the ramp; so there's no motive power on or off from the props.

The total energy TE(0) = KE = 1/2 mV^2 at the foot of the ramp.  At the top of the ramp, TE(h) = ke + we + pe; where ke = 1/2 mv^2, we = kmg cos(theta)s, and pe = mgh.  tan(theta) = h/s; where h = 1.9 m, theta = 28.8 degrees, m = 285 kg, V = 13.5 mps, and k = .15.  s = h/tan(theta); the length of the slide up the ramp's surface

From the conservation of energy TE(0) = 1/2 mV^2 = 1/2 mv^2 + kmg cos(theta)(h/tan(theta)) + mgh = TE(h); where g = 9.81 m/sec^2.  Solve for v = ? which is the velocity off the top of the ramp.  All the values are given, you can do the math.

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

_____ m/s

First, recognize that as long as vy>0 and there is vertical velocity up, the boat will continue to rise to H its max height above ground.  At that point all the KE becomes PE = KE - we = 1/2 mV^2 - kmg cos(theta)s = mgH where we is the work done against friction.  Further PE = pe + mgy; where y is the additional height after leaving the ramp and pe = mgh the potential energy at the top of the ramp.  Thus H = h + y.

The impact velocity u^2 = vx^2 + vy^2 = (v cos(theta))^2 + 2gH); you found v, the top of ramp velocity and you can find H = [1/2 V^2 - kg cos(theta)(h/tan(theta))]/g where everything on the RHS is given.  Plug everything in for u^2 and take the square root to find u = ? the impact velocity.

The physics is simply the conservation of energy throughout the boats travel.  At the bottom of the ramp, all its energy is kinetic KE.  At the top of the ramp its both kinetic ke and potential pe, but some of the KE was also lost we due to friction.  Thus KE = ke + pe + we at the top of the ramp.

Assuming pe + we < KE, that means ke > 0; so there will be additional climb off the ramp until there is no more vertical velocity vy = 0 at height H = h + y.  Further, disregarding drag, vx remains fixed.  Thus, the impact velocity u can be found from u^2 = vx^2 + vy^2; and vx = v cos(theta) and vy^2 = 2gH, where the potential energy mgH is converted to a sort of vertical kinetic energy from the vertical velocity at iimpact.  [You can check your numbers by noting that 1/2 mV^2 = 1/2 mu^2 + we; that is, the initial KE = the impact ke plus the energy lost due to friction.]

The critical part of this energy audit is to ensure you have all the energy accounted for at each waypoint: bottom of ramp, top of ramp, and top of its max height above ground.