Q. A $10kg$ mass moves along $x$-axis. Its acceleration as a function of its position is shown in the figure. What is the total work done on the mass by the force as the mass moves from $x = 0$ to $x = 8\ cm$
A) $8 \times 10^{- 2}\ joules$
B) $16 \times 10^{- 2}\ joules$
C) $4 \times 10^{- 4}\ joules$
D) $1.6 \times 10^{- 3}\ joules$
Explanation
(a) Work done = Area covered in between force displacement curve and displacement axis
= Mass $\times$ Area covered in between acceleration-displacement curve and displacement axis
$= 10 \times \frac{1}{2}\left( 8 \times 10^{- 2} \times 20 \times 10^{- 2} \right) = 8 \times 10^{- 2}J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.