Bob (80kg) takes off a 60m high, 10 degree slope on his jet powered rocket skis.

Question

Bob (80kg) takes off a 60m high, 10 degree slope on his jet powered rocket skis. The coefficients of kinetic and static friction between the skis and the ramp are 0.3 and 0.5. The skis have a thrust of 250N and remain tilted in 10 degrees throughout the problem. While bob is in the air he experiences a drag force of 70N. A) what is the net force acting on bob going up the ramp? B) what is bobs acceleration while on the ramp? C) how far does bob land from the cliff? Bob (80kg) takes off a 60m high, 10 degree slope on his jet powered rocket skis. The coefficients of kinetic and static friction between the skis and the ramp are 0.3 and 0.5. The skis have a thrust of 250N and remain tilted in 10 degrees throughout the problem. While bob is in the air he experiences a drag force of 70N. A) what is the net force acting on bob going up the ramp? B) what is bobs acceleration while on the ramp? C) how far does bob land from the cliff? A) what is the net force acting on bob going up the ramp? B) what is bobs acceleration while on the ramp? C) how far does bob land from the cliff?

Explanation / Answer

here,

mass of bob, m = 80 kg
height of ramp, h = 60m

Distance of slope,d = h/Sin10 = 345.526 m

Coefficient of friction,
us = 0.3
uk = 0.5

Thrust force of skies, Ft = 250 N

Drag force experienced by bob, Fd = 70 N

Part a:
From newton Second Law Sum(F) = m*a
Ft - mgSin10 + Ff - Fd = m*a

Ff = force due to friction

Part B:
Further solving equation form part a
Ft - mgSin10 + Ff - Fd = m*a
Ft - mgSin10 + uk*mgCos10 - Fd = m*a
250 - (80*9.81*Sin10) + (0.5*80*9.81*Cos10) = 80*a

solving for acceleration, a
a = 6.252 m/s^2

Part C:
Distance to fall is given as,
R = v^2*Sin2A/g

Since , v = sqrt(2*a*d) ( From Third eqn of motion)

R = 2*a*d*Sin(2*10)/g
R = 2*6.252*345.526*Sin(2*10)/9.81
R = 150.63 m

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