Question

A force acts on a $2kg$ object so that its position is given as a function of time as $x=3t^2+5$. What is the work done by this force in first $5$ seconds?

• A. $850J$
• B. $875J$
• C. $900J$
• D. $950J$

Answer 1

The correct option is C $900J$

Step1: Find the velocity of the object.
Mass of the object $m=2kg$
Position of the object $x=3t^2+5$
Time $t=5\text{seconds}$
$x=3t^2+5$
$v=\frac{dx}{dt}$
$v=6t+0$
At $t=0$, $v=0$
$t=\text{sec}$
$v=30m/s$

Step2: Find the work done by the force.
Formula Used: $W=\Delta KE$
By work energy theorem
$WD=\Delta KE$
$WD=\frac{1}{2}m{v}^2-0$

$=\frac{1}{2}\left(2\right){\left(30\right)}^2$

$=900J$
Final Answer: (b)