Question

A force acts on a $2\, kg$ 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. 850 J
• B. 900 J
• C. 950 J
• D. 875 J

Answer 1

Answer: (B)

900 J

Answer 2

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