Question

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

• A. $30\text{J}$
• B. $40\text{J}$
• C. $20\text{J}$
• D. $50\text{J}$

Answer 1

The correct option is D $50\text{J}$

Mass of the body $\left(m\right)=0.5\text{kg}$
Velocity of the body $\left(v\right)=a{x}^{3/2}$
where, $a=5{\text{m}}^{-1/2}/\text{s}$
Velocity of the body at $x=0$,
$v_1=5\times 0=0$
Velocity of the body at $x=2\text{m}$,
$v_2=5\times (2{)}^{3/2}\text{m/s}$
According to work-energy theorem,
Work done = Change in kinetic energy
$=\frac{1}{2}m{v}_2^2-\frac{1}{2}m{v}_1^2$
$=\frac{1}{2}m(v_2^2-v_1^2)$
$=\frac{1}{2}\times 0.5\left[\right(5\left(2^{3/2}\right){)}^2-\left(0{)}^2\right]$
$=\frac{1}{2}\times 0.5\times 25\times 2^3$
$=\frac{1}{2}\times 12.5\times 8=50\text{J}$