Q. A box of mass 50 kg is pulled up on an incline 12 m long and 2 m high by a constant force of 100 N from rest. It acquires a velocity of $2\ ms^{- 1}$ on reaching the top. Work done against friction $(g = 10\ ms^{- 2})$ is
A) 50 J
B) 100 J
C) 150 J
D) 200 J
Explanation
(b) If $W_{1} =$ work done by applied force
$W_{2} =$ work done against friction then applying work energy theorem
$W_{1} - W_{2} =$ PE +KE (at the top)
$F \times s - W_{2} = mgh + \frac{1}{2}mv^{2}$
$100 \times 12 - W_{2} = 50 \times 10 \times 2 + \frac{1}{2} \times 50 \times 2^{2}$
$1200 - W_{2} = 1100$
$W_{2} = \ 100\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.