Question

The displacement $x$ (in metre) of a particle of mass $kg$ moving in one dimension under the action of a force is related to the time $t$ (in seconds) by the equation: $t=\sqrt{x}+3.$
The work done in first 6 seconds is:-

• A. $18mJ$
• B. zero
• C. $\frac{9}{2}mJ$
• D. $36mJ$

Answer 1

Answer: (B)

$x=t^2+9-6t$
so $v=2t-6$
Further, At $t=0v=2\times 0-6=-6$
At $t=6v=2\times 6-6=+6$
Initial $KE=\frac{1}{2}m{\left(-6\right)}^2=18m$
Final $KE=\frac{1}{2}m{\left(6\right)}^2=18m$
Work done $=$ Change in $KE$
$W=18m-18m=0$