JEE Mains · Physics · STD 11 - Work,energy,power and collision
- JEE Mains 2023
- Medium
- Numerical
A body of mass \(1 kg\) begins to move under the action of a time dependent force \(\vec{F}=\left(t \hat{i}+3 t^2 \hat{ j }\right) N\). where \(\hat{i}\) and \(\hat{j}\) are the unit vectors along \(x\) and \(y\) axis. The 10.Power developed by above force, at the time \(t=2 s\). will be \(..............W\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(100\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{ F }=t \hat{i}+3 t ^2 \hat{ j }\)
\(\frac{ md \overrightarrow{ v }}{ dt }=t \hat{i}+3 t ^2 \hat{ j }\)
\(m =1 kg , \int \limits_0^{\vec{v}} dv =\int \limits_0^{ t } tdt \hat{ i }+\int \limits_0^{ t } 3 t ^2 dt \hat{ j } ~\\ \overrightarrow{ v }=\frac{ t ^2}{2} \hat{ i }+ t ^3 \hat{ j }\)
\(\text { 10.Power }=\overrightarrow{ F } \cdot \overrightarrow{ V }=\frac{ t ^3}{2}+3 t ^5\) \(\text { At } t =2, \text { 10.Power }=\frac{8}{2}+3 \times 32\)
\(=100\)
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