A body of mass 1 kg begins to… - JEE Main 2023 Physics PYQ

JEE Main 2023 — Physics Mechanics

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A body of mass $1\mathrm{kg}$ begins to move under the action of a time dependent force $\vec{F}=(t\hat{i}+3{t}^{2}\hat{j})N$ , where $\hat{i}$ and $\hat{j}$ are the unit vectors along $x$ and $y$ axis. The power developed by above force, at the time $t=2s$ , will be _______ $W$ .

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Held on 24 Jan 2023 · Verified 6 Jul 2026.

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Solution

Given, $\vec{F}=(t\hat{i}+3{t}^{2}\hat{j})N$ and $m=1\mathrm{kg}$

Now, we can write from Newton's second law, $\vec{F}=m\vec{a}=m\frac{d\vec{v}}{dt}$

Then, $d\vec{v}=(t\hat{i}+3{t}^{2}\hat{j})dt$

Integrating the above,

${\int }_{0}^{v}d\vec{v}={\int }_{0}^{t}(t\hat{i}+3{t}^{2}\hat{j})dt \Rightarrow \vec{v}=\frac{{t}^{2}}{2}\hat{i}+{t}^{3}\hat{j}$

Now, power developed by the force is $P=\vec{F}\cdot \vec{v}=(t\hat{i}+3{t}^{2}\hat{j})\cdot (\frac{{t}^{2}}{2}\hat{i}+{t}^{3}\hat{j})$

$=(\frac{{t}^{3}}{2}+3{t}^{5})$

At $t=2s$ , power is $P=(\frac{{2}^{3}}{2}+3{(2)}^{5})=4+(3\times 32)=100W$

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Solution

Given, $\vec{F}=(t\hat{i}+3{t}^{2}\hat{j})N$ and $m=1\mathrm{kg}$

Now, we can write from Newton's second law, $\vec{F}=m\vec{a}=m\frac{d\vec{v}}{dt}$

Then, $d\vec{v}=(t\hat{i}+3{t}^{2}\hat{j})dt$

Integrating the above,

${\int }_{0}^{v}d\vec{v}={\int }_{0}^{t}(t\hat{i}+3{t}^{2}\hat{j})dt \Rightarrow \vec{v}=\frac{{t}^{2}}{2}\hat{i}+{t}^{3}\hat{j}$

Now, power developed by the force is $P=\vec{F}\cdot \vec{v}=(t\hat{i}+3{t}^{2}\hat{j})\cdot (\frac{{t}^{2}}{2}\hat{i}+{t}^{3}\hat{j})$

$=(\frac{{t}^{3}}{2}+3{t}^{5})$

At $t=2s$ , power is $P=(\frac{{2}^{3}}{2}+3{(2)}^{5})=4+(3\times 32)=100W$

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Solution

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Solution

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JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

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