MCQ
A force acts on a $2\,kg$ object so that its position is given as a function of time as $x= 3t^2 + 5.$ What is the work done by this force in first $5\,seconds$ ? ................ $\mathrm{J}$
- A$850$
- B$950$
- C$875$
- ✓$900$
✓
Answer
Correct option: D.
$900$
d
$\begin{array}{l}
x = 3{t^2} + 5\\
\Rightarrow v = 6t\\
\Rightarrow \,\Delta W = \Delta k\\
\,\,\,\,\,\,\,\,\, = \frac{1}{2}\left( 2 \right){\left( {30} \right)^2} - \frac{1}{2}2{\left( 0 \right)^2}\\
\,\,\,\,\,\,\,\,\, = 900\,J
\end{array}$
$\begin{array}{l}
x = 3{t^2} + 5\\
\Rightarrow v = 6t\\
\Rightarrow \,\Delta W = \Delta k\\
\,\,\,\,\,\,\,\,\, = \frac{1}{2}\left( 2 \right){\left( {30} \right)^2} - \frac{1}{2}2{\left( 0 \right)^2}\\
\,\,\,\,\,\,\,\,\, = 900\,J
\end{array}$
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