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. $900J$
• C. $950J$
• D. $875J$

Answer 1

Answer: (B)

$900J$

$x=3t^2+5$
$v=\frac{dx}{dt}$
$v=6t+0$
$att=0$ $v=0$
t = 5 sec v = 30 m/s
W.D. = $\Delta KE$
$W.D.=\frac{1}{2}m{v}^2-0=\frac{1}{2}\left(2\right)(30{)}^2=900J$