JEE Main 2016 · 9 Apr · Shift 1 · Physics Q4

← JEE Main 2016 · 9 Apr · Shift 1

PhysicsSingle correct+4 / -1Question 4 of 90

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 $\dfrac W{20}$ 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 $\dfrac P2$ while moving downhill at speed $v$ then value of $v$ is :

20 ms

^{-1}

15 ms

^{-1}

10 ms

^{-1}

5 ms

^{-1}

Answer and worked solution

Official answer: B

The slope gives $\sin\theta=\frac{100}{1000}=0.1$ , so gravity along the road is $\frac W{10}$ . Uphill the car must supply $\frac W{20}+\frac W{10}=\frac{3W}{20}$ , so $P=\frac{3W}{20}\times10=1.5W$ .

Downhill the net force the engine must balance is $\left|\frac W{20}-\frac W{10}\right|=\frac W{20}$ (here gravity exceeds friction, so strictly the car would need braking; the intended reading uses this magnitude). Then $\frac P2=\frac W{20}v$ gives $v=15$ m s⁻¹.

Topics:Power Friction

← Previous question Next question →

JEE Main 2016 · 9 Apr · Shift 1, Physics question 4