A body of mass $0.5 \mathrm{~kg}$ travels on straight line path with velocity $v=\left(3 x^{2}+4\right) \mathrm{m} / \mathrm{s}$. The net workdone by the force during its displacement from $x=0$ to $x=2 \mathrm{~m}$ is :
Solution
<p>$v = 3{x^2} + 4$</p>
<p>at $x = 0$, ${v_1} = 4$ m/s</p>
<p>$x = 2$, ${v_2} = 16$ m/s</p>
<p>$\Rightarrow$ Work done = $\Delta$ kinetic energy</p>
<p>$= {1 \over 2} \times m\left( {v_2^2 - v_1^2} \right)$</p>
<p>$= {1 \over 4}(256 - 16)$</p>
<p>$= 60$ J</p>
About this question
Subject: Physics · Chapter: Work, Energy and Power · Topic: Work Done by a Force
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