Q. A car of weight $W$ is on an inclined road that rises by $100\,m$ over a distance of $1\,km$ 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 :
Solution:
$P=\left(w\,sin\,\theta+\frac{w}{20}\right)10$
$P=\left(\frac{w}{10}+\frac{w}{20}\right)10=\frac{3w}{2}$
$\frac{P}{2}=\frac{3w}{4}=\left(\frac{w}{10}-\frac{w}{20}\right)w$
$\frac{3}{4}=\frac{v}{20}$
$15m/s=v$
