7. (a) Figure 1 shows a cyclist going up a hill.
Figure 1
The angle $\theta$ of the slope of the hill is constant.
The total mass $m$ of the cyclist and bicycle is $65 \mathrm{~kg}$.
Write an expression for the component of the total weight parallel to the slope.
$65 m g \sin \theta$
(b) The useful power output of the cyclist is $310 \mathrm{~W}$.
The cyclist has a steady speed of $1.63 \mathrm{~m} \mathrm{~s}^{-1}$.
Assume that air resistance is negligible at this speed.
Calculate $\theta$.
(1)
Calculate $\theta$
\[
\begin{array}{l}
p=f V \\
\frac{310}{1.63}=190.1840491
\end{array}
\]
\[
\theta=
\]
(5)

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 \(65mg\sin\theta\). 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): \[F = \frac{P}{V}\]

Step 3 ：Substitute the given values: \[F = \frac{310}{1.63} = 190.1840491\]

Step 4 ：Now that we have the force (F) acting on the cyclist parallel to the slope, we can use the equation \(F = 65mg\sin\theta\) 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: \[\sin\theta = \frac{F}{65mg}\]

Step 5 ：Substitute the values and solve for θ: \[\theta = \arcsin\left(\frac{190.1840491}{65 \times 9.81}\right) \approx 17.35°\]

Step 6 ：\(\boxed{\theta \approx 17.35°}\)