Question

A particle of mass m = 10-2 kg is moving along thepositive x-axis under the influence of a force,F=-k/2 x2, where k = 10-2 Nm2. When the particleis at x = 1.0 m, its velocity v = 0. When it reachesx = 0.5 m, the magnitude of its velocity is(a) 0.5 m s-1(b) 1.0 ms-1(c) 1.5 m s-1(d) 2.0 ms-1​

Answer 1

answer : option (b) 1.0 m/s²

explanation : given, F = -k/2x²

from Newton's 2nd law, F = ma

we know, a = v dv/dx

so, F = mv dv/dx

⇒F = mv dv/dx = -k/2x²

⇒∫vdv = -k/2m ∫dx/x²

⇒v²/2 = -k/2m [-1/x] + C , where C is constant of integration.

⇒v²/2 = k/2mx + C

given, at x = 1, v = 0

so, 0²/2 = k/2m(1) + C

or, C = -k/2m

now, v²/2 = k/2mx - k/2m

⇒ v² = k/m [1/x - 1]

putting values of k, m and x

then, v² = (10^-2)/(10^-2) [1/0.5 - 1] = 1

or, v = 1

hence, magnitude of velocity at x = 0.5 is 1.0 m/s

Answer 2

Answer: The particle is moving against the action of force. Therefore the magnitude of velocity will be 1m/s.