Asked by lost
sorry here is actual 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 28.5° 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 1.70 m above the water surface.
a. speed of boat when leaves ramp (engine off once hits ramp)
b. speed of boat when hits water again (no air resistance)
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 28.5° 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 1.70 m above the water surface.
a. speed of boat when leaves ramp (engine off once hits ramp)
b. speed of boat when hits water again (no air resistance)
Answers
Answered by
Elena
The work-energy theorem
KE2 = KE1 –W(fr) –PE,
KE2 = m•v1²/2 - k•m•g•cosα•s - m•g•h =
= m•v²/2 - m•g•h{(k/tanα) +1} = 2•10^4 J.
KE2 = m•v2²/2,
v2 = sqrt(2•KE2/m) = 11.8 m/s,
The work-kinetic energy theorem for the
boat while it is airborne:
KE3 = ΔPE+KE2 = m•g•h+ m•v2²/2 = 2.46•10^4 J.
KE3 = m•v3²/2,
v3 = sqrt(2•KE3/m) = 13.17 m/s.
KE2 = KE1 –W(fr) –PE,
KE2 = m•v1²/2 - k•m•g•cosα•s - m•g•h =
= m•v²/2 - m•g•h{(k/tanα) +1} = 2•10^4 J.
KE2 = m•v2²/2,
v2 = sqrt(2•KE2/m) = 11.8 m/s,
The work-kinetic energy theorem for the
boat while it is airborne:
KE3 = ΔPE+KE2 = m•g•h+ m•v2²/2 = 2.46•10^4 J.
KE3 = m•v3²/2,
v3 = sqrt(2•KE3/m) = 13.17 m/s.
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