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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)


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}


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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