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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 :
Option: 1
20 ms⁻¹

Option: 2
15 ms⁻¹

Option: 3
10 ms⁻¹

Option: 4
5 ms⁻¹

Answers (1)

best_answer

From the question

$\sin \Theta=\frac{100}{1000} = \frac{1}{10}$

where $\Theta$ = angle made by incline plane with the horizontal

While going up,

the net force on car= $Wsin\Theta +f=\frac{W}{10}+\frac{W}{20}$

Power=P= $F.V$

using V=10 m/s

i.e $\frac{3W}{2} =P$

When going down,

the net force on car= $Wsin\Theta -f=\frac{W}{10}-\frac{W}{20}=\frac{W}{20}$

Power= $=\frac{P}{2} =\frac{W}{20}*V_2$

On comparing we get $V_2=15 \ m/s$

Posted by

Divya Prakash Singh

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