1. Tardigrade
  2. Question
  3. Physics
  4. A force acts on a 2 kg object so that its position is given as a function of time as x = 3t2 + 5. What is the work done by this force in first 5 seconds ?

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. 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 ?

JEE MainJEE Main 2019Work, Energy and Power

Solution:

$x = 3t^2 + 5$
$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 $