Question

A car of weight $W$ is on 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 $10m{s}^{-1}$, the car needs power $P$. If it needs power $\frac{P}{2}$ while moving downhill at speed $v$ then value of $v$ is :

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

Answer 1

The correct option is D $5m{s}^{-1}$

We know that if $inclination$ from horizontal is $\theta$ then $Sin\theta =\frac{100m}{1km}=\frac{100}{1000}=0.1$

So gravity force on car $down$ the inclination will be $mgSin\theta =Wsin\theta =0.1W$

and $up$ the incline $\left(supporting\right)$ friction is $F_r=0.05W$ so force needed by car to move with $constant$ velocity will be

$F=0.1W-F_r=0.1W-0.05W=0.05W$

so power will be $P=Fv=0.05W\times 10m/s=0.5Wwatt$

When it will move down the hill then $friction$ will be $up$ the incline and force by $gravity$ will be $down$ the incline so

net force will be down the incline, in this case the car need to apply $brake$ to make it move with $constant$ velocity

so force needed by $brakes$ will be $F_0=0.1W-F_r=0.1W-0.05W=0.05W$

thus we get new power as $P_0=F_{0}v=0.05W\times v$

but it should be like $P_0=\frac{P}{2}$ so $0.05W\times v=\frac{0.5W}{2}$

so $v=5m/s$