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Step-by-Step Solution
Step 1: Understand the Problem
We have a car of weight $W$ traveling on a road inclined such that it rises by 100 m over a 1000 m distance. The car experiences a constant frictional force of $W/20.$ We are given:
Uphill speed $u = 10\,\text{m/s}$.
Power needed uphill = $P$.
Power needed downhill = $P/2$ at some speed $v$ (to be found).
We must find the speed $v$ when the car travels downhill, given the same incline and friction.
Step 2: Determine the Inclination Angle
The road rises 100 m vertically over a distance of 1000 m. Thus,
$$
\tan\theta = \frac{100}{1000} = \frac{1}{10}.
$$
Since $\theta$ is quite small, we approximate
$$
\sin \theta \approx \frac{1}{10}.
$$
Step 3: Express the Power Required Uphill
When going uphill, two opposing forces must be overcome: the component of weight along the incline ($W \sin\theta$) and the frictional force ($W/20$). Therefore, the total resisting force is
$$
F_\text{uphill} = W \sin\theta + \frac{W}{20}.
$$
Given the speed uphill is $u=10\,\text{m/s}$, power is force multiplied by velocity:
$$
P = F_\text{uphill} \times u.
$$
Substitute $ \sin\theta = \frac{1}{10} $ and $ u = 10\,\text{m/s}$:
$$
P = \left(\frac{W}{10} + \frac{W}{20}\right) \times 10.
$$
Simplify:
$$
P = \left(\frac{2W}{20} + \frac{W}{20}\right) \times 10
= \left(\frac{3W}{20}\right) \times 10
= \frac{3W}{2}.
$$
Step 4: Express the Power Required Downhill
When going downhill, the component of weight along the incline ($W \sin\theta$) now aids the motion, but friction ($W/20$) still opposes it. Hence, the net force to be overcome by the engine is
$$
F_\text{downhill} = W \sin\theta - \frac{W}{20}.
$$
We know the power required downhill is half of $P$, which is $P/2$. Denote the downhill speed as $v$. Then,
$$
\frac{P}{2} = F_\text{downhill} \times v.
$$
We have
$$
\frac{P}{2} = \left( W \sin\theta - \frac{W}{20} \right) v.
$$
Substitute $ P = \frac{3W}{2} $ and $ \sin\theta = \frac{1}{10} $:
$$
\frac{\frac{3W}{2}}{2}
= \left( \frac{W}{10} - \frac{W}{20} \right) v.
$$
This simplifies to
$$
\frac{3W}{4}
= \left( \frac{W}{10} - \frac{W}{20} \right) v.
$$
Step 5: Solve for the Downhill Speed $v$
Inside the parenthesis, factor out $W$:
$$
\frac{3W}{4} = \left(\frac{W}{10} - \frac{W}{20}\right) v
= \left(\frac{2W}{20} - \frac{W}{20}\right) v
= \frac{W}{20} \, v.
$$
Cancel $W$ from both sides (assuming $W \neq 0$):
$$
\frac{3}{4} = \frac{v}{20}.
$$
Hence,
$$
v = 20 \times \frac{3}{4} = 15\,\text{m/s}.
$$
Step 6: Final Answer
The downhill speed $v$ at which the car needs power $P/2$ is
$$
\boxed{15\,\text{m/s}}.
$$
Reference Images
Please refer to the figures provided for a visual representation of the uphill and downhill motion.
