Question

A bike and cyclist have a total mass of 80kg and are moving at a constant initial speed of 10.m⋅s−1. It takes the cyclist 1.0 minute to ride up a steep hill at a constant speed of 10.m⋅s−1. The hill is 30.m high. She reaches the top and then horizontally coasts to a stop in 5s. Assuming all resistive forces are constant, what is the power of the cyclist as she climbs the hill? Give the answer in kW correct to one significant figure and do not include units in your answer

Answer #1

Time travelled horizontally before stopping = 5s

initial speed = 10 m/s

=> acceleration = 10/5 = 2 m/s^{2}

=>Coefficient of friction u = a/g = 0.2

Time taken to reach top of hill = 1 minute = 60s

Speed = 10 m/s

=> Distance travelled = 10*60 = 600 m/s

Slope Θ = sin^{-1} ( 30 / 600 ) = 2.87 degrees

Power output of cyclist = dW/dt = d(Fs)/dt = F ds/ dt = F * v

Now, force on cyclist along slope downwards due to gravity
F_{g}= m * g * sin ( Θ ) = 80*10*(30/600) = 40 N

Force due to friction opposing the motion F_{f} = u
*normal force = u m g cos ( Θ ) = 0.2 * 80 *10 * cos ( 2.87 ) =
159.7998 N

So total force F = F_{g} + F_{f} = 40 + 159.7998
= 199.7998 N

**Power = F v = 199.7998 * 10 = 1997.998 W = 1.998
kW**

**Ans = 1.998**

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