Q. A car of mass $1250\ kg$ is moving at $30\ m/s$. Its engine delivers $30\ kW$ while resistive force due to surface is $750\ N$. What max acceleration can be given in the car
A) $\frac{1}{3}\ m/s^{2}$
B) $\frac{1}{4}\ m/s^{2}$
C) $\frac{1}{5}\ m/s^{2}$
D) $\frac{1}{6}\ m/s^{2}$
Explanation
(c) Force produced by the engine $F = \frac{P}{v} = \frac{30 \times 10^{3}}{30} = 10^{3}N$
Acceleration = $\frac{Forward\ force\ by\ engine\ - \ resistive\ force}{mas\ of\ car}$
$= \frac{1000 - 750}{1250} = \frac{250}{1250} = \frac{1}{5}\ m/s^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.