Under the action of a force, a 2 ,kg body mov | vedclass

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(A) Given: $m = 2 \,kg$, $x = \frac{\alpha t^2}{2}$, and $\alpha = 1 \,m/s^2$.First, we find the velocity $v$ by differentiating the position $x$ with

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$4$

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(A) Given: $m = 2 \,kg$, $x = \frac{\alpha t^2}{2}$, and $\alpha = 1 \,m/s^2$.First, we find the velocity $v$ by differentiating the position $x$ with respect to time $t$:$v = \frac{dx}{dt} = \frac{d}{dt} \left( \frac{\alpha t^2}{2} ight) = \alpha t$.At $t = 2 \,s$, the velocity $v$ is:$v = 1 	imes 2 = 2 \,m/s$.According to the work-energy theorem, the work done $W$ is equal to the change in kinetic energy:$W = \Delta K = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$.At $t = 0$, $v_i = \alpha(0) = 0$.At $t = 2 \,s$, $v_f = 2 \,m/s$.Therefore, $W = \frac{1}{2} 	imes 2 \,kg 	imes (2 \,m/s)^2 - 0 = 4 \,J$.

Under the action of a force, a $2 \,kg$ body moves such that its position $x$ as a function of time $t$ is given by $x = \alpha t^2 / 2$, where $x$ is in meters, $t$ is in seconds, and $\alpha = 1 \,m/s^2$. The work done by the force in the first two seconds is: (in $\,J$)

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