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A body of mass $5 \mathrm{~kg}$ is moving in a straight line. The relation between its
displacement and time is $\mathrm{x}=\left(\mathrm{t}^{3}-2 \mathrm{t}-10\right) \mathrm{m}$. What is the force acting on it at the end of 5 second?
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displacement and time is $\mathrm{x}=\left(\mathrm{t}^{3}-2 \mathrm{t}-10\right) \mathrm{m}$. What is the force acting on it at the end of 5 second?
Solution:
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Verified Answer
The correct answer is:
$150 \mathrm{~N}$
A body of mass rig,
\begin{array}{l}
m=5 k g \\
x=\left(t^{3}-2 t-10\right) m
\end{array}
F after $t=5 \mathrm{sec}$,
\begin{array}{l}
\frac{d x}{d t}=v=3 t^{2}-2 \\
\frac{d v}{d t}=a=6 t \\
F=m a=5 \times 6 t=30 \times 5 \\
F=150 \mathrm{~N}
\end{array}
\begin{array}{l}
m=5 k g \\
x=\left(t^{3}-2 t-10\right) m
\end{array}
F after $t=5 \mathrm{sec}$,
\begin{array}{l}
\frac{d x}{d t}=v=3 t^{2}-2 \\
\frac{d v}{d t}=a=6 t \\
F=m a=5 \times 6 t=30 \times 5 \\
F=150 \mathrm{~N}
\end{array}
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