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

Question

A 285-kg stunt boat is driven on the surface of a lake at aconstant speed of 13.5 m/s toward a ramp, which is angled at27.1° above the horizontal. Thecoefficient of friction between the boat bottom and the ramp'ssurface 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 theboat hits the ramp, what is the speed of the boat as it leaves theramp?
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 (a) Assuming the engines are cut off when theboat hits the ramp, what is the speed of the boat as it leaves theramp?
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) 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; sothere'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. Atthe 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 = 27.1 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 + kmgcos(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. Allthe 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 verticalvelocity up, the boat will continue to rise to H its max heightabove ground. At that point all the KE becomes PE = KE - we = 1/2mV^2 - kmg cos(theta)s = mgH where we is the work done againstfriction. Further PE = pe + mgy; where y is the additional heightafter leaving the ramp and pe = mgh the potential energy at the topof 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 isgiven. Plug everything in for u^2 and take the square root to findu = ? the impact velocity. The physics is simply the conservation of energy throughout theboats travel. At the bottom of the ramp, all its energy is kineticKE. 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 0; so there will beadditional climb off the ramp until there is no more verticalvelocity vy = 0 at height H = h + y. Further, disregarding drag, vxremains 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 thepotential energy mgH is converted to a sort of vertical kineticenergy from the vertical velocity at iimpact. [You can check yournumbers by noting that 1/2 mV^2 = 1/2 mu^2 + we; that is, theinitial KE = the impact ke plus the energy lost due tofriction.] The critical part of this energy audit is to ensure you have allthe energy accounted for at each waypoint: bottom of ramp, top oframp, and top of its max height above ground

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