JEE MAIN - Physics (2022 - 28th June Evening Shift - No. 3)
A block of mass 2 kg moving on a horizontal surface with speed of 4 ms$$-$$1 enters a rough surface ranging from x = 0.5 m to x = 1.5 m. The retarding force in this range of rough surface is related to distance by F = $$-$$kx where k = 12 Nm$$-$$1. The speed of the block as it just crosses the rough surface will be :
zero
1.5 ms$$-$$1
2.0 ms$$-$$1
2.5 ms$$-$$1
Explanation
$$F = - 12x$$
$$mv{{dv} \over {dx}} = - 12x$$
$$\int_4^v {vdv = - 6\int_{0.5}^{1.5} {xdx} } $$ ($$m = 2$$ kg)
$${{{v^2} - 16} \over 2} = - 6\left[ {{{{{1.5}^2} - {{0.5}^2}} \over 2}} \right]$$
$${{{v^2} - 16} \over 2} = - 6$$
$$v = 2$$ m/sec
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