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A block of mass $2 \mathrm{~kg} F(t) \uparrow$ is free to move along the $x$-axis. It is at rest and from $t=0$ onwards it is subjected to a timedependent force $F(t)$ in the $x$ direction. The force $F(t)$ varies with $t$ as shown in the figure. The kinetic energy of the block after 4.5 seconds is
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The correct answer is:
$5.06 \mathrm{~J}$
Change in momentum $=$ area under $F$ - $t$ graph
$\begin{aligned} \Delta p & =\frac{1}{2} \times 4 \times 3-\frac{1}{2} \times 1.5 \times \frac{4}{3} \times 1.5 \\ & {\left[\text { Force at } 4.5 \mathrm{~s}=\frac{4}{3} \times 1.5 \mathrm{~N}\right] } \\ & =6-1.5=4.5 \mathrm{~kg} \mathrm{~m} \mathrm{~s}\end{aligned}$
As $\Delta p=m(v-u)=m v \quad(\because u=0)$
$\therefore \quad v=\frac{\Delta p}{m}=\frac{4.5 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}}{2 \mathrm{~kg}}=\frac{4.5}{2} \mathrm{~ms}^{-1}$
Kinetic energy of the block after $4.5 \mathrm{~s}$ is
$K=\frac{1}{2} m v^2=\frac{1}{2} \times 2 \mathrm{~kg} \times\left(\frac{4.5}{2} \mathrm{~ms}^{-1}\right)^2=5.06 \mathrm{~J}$
$\begin{aligned} \Delta p & =\frac{1}{2} \times 4 \times 3-\frac{1}{2} \times 1.5 \times \frac{4}{3} \times 1.5 \\ & {\left[\text { Force at } 4.5 \mathrm{~s}=\frac{4}{3} \times 1.5 \mathrm{~N}\right] } \\ & =6-1.5=4.5 \mathrm{~kg} \mathrm{~m} \mathrm{~s}\end{aligned}$
As $\Delta p=m(v-u)=m v \quad(\because u=0)$
$\therefore \quad v=\frac{\Delta p}{m}=\frac{4.5 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}}{2 \mathrm{~kg}}=\frac{4.5}{2} \mathrm{~ms}^{-1}$
Kinetic energy of the block after $4.5 \mathrm{~s}$ is
$K=\frac{1}{2} m v^2=\frac{1}{2} \times 2 \mathrm{~kg} \times\left(\frac{4.5}{2} \mathrm{~ms}^{-1}\right)^2=5.06 \mathrm{~J}$
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