Question

A particle of mass 10^-2 kg is moving along the positive x-axis under the influence of a force F(x) = `-"K"/(2x)^2` where K = 10^-2 Nm².  At time t = 0 it is at x = 1.0 m and its velocity is v = 0. The velocity of particle will be ______ m/s, when it reaches x = 0.50 m.

पर्याय

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Answer 1

A particle of mass 10^-2 kg is moving along the positive x-axis under the influence of a force F(x) = `-"K"/(2x)^2` where K = 10^-2 Nm².  At time t = 0 it is at x = 1.0 m and its velocity is v = 0. The velocity of particle will be 1 m/s, when it reaches x = 0.50 m.

Explanation:

F = `(-"k")/(2x^2)`

⇒ F = `-(10^-2)/(2x^2)`

a = `"F"/"m" = 10^-2/((10^-2)2x^2)`

⇒ a = `(-1)/(2x^2) = "v" "dv"/"dx"`

`-int_1^0.5("d"x)/(2x^2) = int_0^"v""vdv"`

⇒ `-1/2int_1^0.5x^-2dx="v"^2/2`

⇒ `-1/2[-1/x]_1^0.5 = "v"^2/2`

⇒ `1/2[1/0.5-1]="v"^2/2`

⇒ v = 1 m/s