Question

A body of mass 0.5 kg travels in a straight line with velocity $v=a{x}^{3/2}$ where $a=5m^{-1/2}{s}^{-1}$. The work done by the net force during its displacement from x = 0 to x = 2 m is

• A. 1.5 J
• B. 50 J
• C. 10 J
• D. 100 J

Answer 1

Answer: (B)

50 J

Velocity of body is given as $v=a{x}^{3/2}$

where $a=5$

Initial velocity of body $v_i{|}_{x=0}=5\times 0=0m/s$

Final velocity of body $v_f{|}_{x=2}=5\times 2^{3/2}=10\sqrt{2}m/s$

From work-energy theorem, work done by the force is equal to change in kinetic energy of the body.

$\therefore$ $W=\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2$

Or $W=\frac{1}{2}\left(0.5\right)(10\sqrt{2}{)}^2-\frac{1}{2}(0.5\left)\right(0)$

Or $W=\frac{1}{2}\left(0.5\right)\left(200\right)-0$

$⟹W=50J$