JEE MAIN - Physics (2022 - 25th July Morning Shift - No. 4)
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 :
64 J
60 J
120 J
128 J
Explanation
$$v = 3{x^2} + 4$$
at $$x = 0$$, $${v_1} = 4$$ m/s
$$x = 2$$, $${v_2} = 16$$ m/s
$$\Rightarrow$$ Work done = $$\Delta$$ kinetic energy
$$ = {1 \over 2} \times m\left( {v_2^2 - v_1^2} \right)$$
$$ = {1 \over 4}(256 - 16)$$
$$ = 60$$ J
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