Question

A car of weight $W$ is on an inclined road that rises by $100m$ over a distance of $1km$ and applies a constant frictional force $\frac{W}{20}$ on the car. While moving uphill on the road at a speed of $10{ms}^{-1}$, the car needs power $P$. If it needs power $\frac{P}{2}$ while moving downhill at speed $v$ then value of $v$ is:

• A. $15{ms}^{-1}$
• B. $5{ms}^{-1}$
• C. $10{ms}^{-1}$
• D. $20{ms}^{-1}$

Answer 1

The correct option is A $15{ms}^{-1}$

Let us apply work energy theorem in this question. In the first case,

$100W+\frac{W}{20}\times 1000=Pt=P\times \left(\frac{1000}{10}\right)$

$P=1.5W$

In the downhill case, gravity will support the motion, hence,

$100W-50W=\frac{P{t}^{\prime }}{2}=\frac{1.5W{t}^{\prime }}{2}$

$t^{\prime }=\frac{1000}{15}s$

Hence, $v=\frac{1000}{\frac{1000}{15}}=15m/s$