JEE MAIN - Physics (2022 - 29th June Morning Shift - No. 7)
A particle of mass 500 gm is moving in a straight line with velocity v = b x5/2. The work done by the net force during its displacement from x = 0 to x = 4 m is : (Take b = 0.25 m$$-$$3/2 s$$-$$1).
2 J
4 J
8 J
16 J
Explanation
$${W_{total}} = \Delta K$$
$$ = {1 \over 2}\left( {{1 \over 2}} \right)\left[ {{{\{ b{{(4)}^{5/2}}\} }^2} - 0} \right]$$
$$ = {{{b^2}} \over 4} \times {4^5}$$
$$ \Rightarrow {W_{total}} = 16\,J$$
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