JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
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 \(875\)
- D \(900\)
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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