Question

The displacement $x$ (in meter) of a particle of mass $m\text{(in kg)}$ varies with time $t$ (in seconds) as given by the equation $x=(t-3{)}^2$. Net work done on the particle (in $J$) in first six seconds is

• A. $\left(12m\right)\text{J}$
• B. Zero
• C. $\left(\frac{9}{2}\right)m\text{J}$
• D. $\left(36m\right)\text{J}$

Answer 1

The correct option is B Zero

The displacement is given by $x=(t-3{)}^2=t^2-6t+9$
$v=\frac{dx}{dt}=2t-6$,
we know that velocity is the time derivative of displacement.

at $t=0,v_1=2\times 0-6=-6$
at $t=6,v_2=2\times 6-6=6$

$K_1=\frac{1}{2}m{v}_1^2=\frac{1}{2}m\times 36=10m$

$K_2=\frac{1}{2}m{v}_2^2=\frac{1}{2}m\times 36=10m$

$\Delta K=K_2-K_1=0$

From Work energy theorem, $W=\Delta K$
Hence no work is done.

Hence option B is the correct answer