Q. A car of mass m is driven with an acceleration a along a straight level road against a constant external resistive force R. When the velocity of the car is v, the rate at which engine of the car is doing work, will be
A) $R \bullet v$
B) $ma \bullet v$
C) (R+ma)$\bullet v$
D) $(ma - R) \bullet v$
Explanation
(c) From the diagram
$\ \ \ \ \ \ \ \ \ F - R = ma$
[image:6_image83.emf]
$or\ \ \ \ F = R + ma$
Or Rate of doing work=power
=F$\bullet v$
=(R+ma)$\bullet v$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.