Problem

7. (a) Figure 1 shows a cyclist going up a hill.
Figure 1
The angle θ of the slope of the hill is constant.
The total mass m of the cyclist and bicycle is 65 kg.
Write an expression for the component of the total weight parallel to the slope.
65mgsinθ
(b) The useful power output of the cyclist is 310 W.
The cyclist has a steady speed of 1.63 m s1.
Assume that air resistance is negligible at this speed.
Calculate θ.
(1)
Calculate θ
p=fV3101.63=190.1840491
θ=
(5)

Answer

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Answer

θ17.35°

Steps

Step 1 :Given the power output (P) of the cyclist as 310 W, the total mass (m) as 65 kg, and the steady speed (V) as 1.63 m/s. We need to find the angle θ of the slope. We know that the component of the total weight parallel to the slope is given by 65mgsinθ. We also know that power (P) is equal to force (F) multiplied by velocity (V).

Step 2 :First, we need to find the force (F) acting on the cyclist parallel to the slope. We can do this by dividing the power (P) by the velocity (V): F=PV

Step 3 :Substitute the given values: F=3101.63=190.1840491

Step 4 :Now that we have the force (F) acting on the cyclist parallel to the slope, we can use the equation F=65mgsinθ to find θ. We know the mass (m) is 65 kg, and the gravitational acceleration (g) is approximately 9.81 m/s². We can rearrange the equation to solve for θ: sinθ=F65mg

Step 5 :Substitute the values and solve for θ: θ=arcsin(190.184049165×9.81)17.35°

Step 6 :θ17.35°

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