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JEE Mains · Physics · STD 11 - Work,energy,power and collision
  • JEE Mains 2019
  • Medium
  • Numerical

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 \(5\,seconds\) ? ................ \(\mathrm{J}\)

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Answer & Solution

Correct Answer

(D) \(900\)

Step-by-step Solution

Detailed explanation

\(\begin{array}{l}
x = 3{t^2} + 5\\
 \Rightarrow v = 6t\\
 \Rightarrow \,\Delta W = \Delta k\\
\,\,\,\,\,\,\,\,\, = \frac{1}{2}\left( 2 \right){\left( {30} \right)^2} - \frac{1}{2}2{\left( 0 \right)^2}\\
\,\,\,\,\,\,\,\,\, = 900\,J
\end{array}\)
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