A force acts on a $30\ g$ particle in such a way that the position of the particle as a function of time is given by $x = 3t - 4t^{2} + t^{3}$, where $x$ is in $metres$ and $t$ is in seconds. The work done during the first 4 seconds is

Question

Dr. Abubacker · Accepted Answer

(a) $v = \frac{dx}{dt} = 3 - 8t + 3t^{2}$
$	herefore v_{0} = 3\ m/s$ and $v_{4} = 19\ m/s$
$W = \frac{1}{2}m(v_{4}^{2} - v_{0}^{2})$ [According to work energy theorem]
$= \frac{1}{2} 	imes 0.03 	imes \left( 19^{2} - 3^{2} ight) = 5.28\ J$

Q. A force acts on a $30\ g$ particle in such a way that the position of the particle as a function of time is given by $x = 3t - 4t^{2} + t^{3}$, where $x$ is in $metres$ and $t$ is in seconds. The work done during the first 4 seconds is

Explanation

Reviewed by Dr. Abubacker

Want more practice?