Question
The relation between the position and time of a particle under the influence of constant force is $x$ $=(t-4)^2$, where $x$ is in meters and $t$ is in seconds. Then calculate the work done by force in the first 8 seconds.
✓
Answer
From the work energy theorem,
$\begin{aligned}
\text { work done } & =\text { change in kinetic energy } \\
& =\left(\frac{1}{2} m v_{-}^2-\frac{1}{2} m v_{\# 0}^2\right)
\end{aligned}$
Here $x=(t-4)^2$
Velocity $v =\frac{d x}{d t}=\frac{d}{d t}(t-4)^2$
$v=2(t-4)$
Hence,
$\begin{aligned}
v_{t=0} & =2(0-4)=-8 m / s \\
v_{t=8} & =2(8-4)=8 m / s \\
W & =\frac{1}{2} m\left[(8)^2-(-8)^2\right] \\
& =\frac{1}{2} m[64-64]=0
\end{aligned}$
or $W=0\quad \text { Ans. }$
$\begin{aligned}
\text { work done } & =\text { change in kinetic energy } \\
& =\left(\frac{1}{2} m v_{-}^2-\frac{1}{2} m v_{\# 0}^2\right)
\end{aligned}$
Here $x=(t-4)^2$
Velocity $v =\frac{d x}{d t}=\frac{d}{d t}(t-4)^2$
$v=2(t-4)$
Hence,
$\begin{aligned}
v_{t=0} & =2(0-4)=-8 m / s \\
v_{t=8} & =2(8-4)=8 m / s \\
W & =\frac{1}{2} m\left[(8)^2-(-8)^2\right] \\
& =\frac{1}{2} m[64-64]=0
\end{aligned}$
or $W=0\quad \text { Ans. }$
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