Q. A car weighing 1400 kg is moving at a speed of $54\ kmh^{- 1}$ up a hill when the motor stops. If it is just able to reach the destination which is at a height of 10 m above the point, then the work done against friction (negative of the work done by the friction) is [Take $g = 10ms^{- 2}$]
A) 10 kJ
B) 15 kJ
C) 17.5 kJ
D) 25 kJ
Explanation
(c) $1400 \times 10 \times 10 + W = \frac{1}{2} \times 15 \times 15$
or $W = 700 \times 15 \times 15 - 1400 \times 10 \times 10$
or $W = 700(225 - 200)\ J$
or $W = 700 \times 25\ J\ = 75.5\ kJ$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.