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# A car of mass m is driven with an acceleration 'a' along a straight level road. against a constant external resistive force R. The velocity of the car is V, the rate at which the engine of the car is doing work is

$(a)\;RV\quad (b)\;(R+ma)V \quad (c)\;maV \quad (d)\;(ma+V)R$

Rate of work done$=\large\frac{Force \times displacement}{time}$
$\quad= Force \times velocity$
Total force $=R+ma$
Therefore Rate of work done $=(R+ma)V$
Hence b is the correct answer.

edited Feb 10, 2014 by meena.p