Question

Two blocks of equal mass m are connected by an unstretched spring and the system is kept at rest on a frictionless horizontal surface. A constant force F is applied on one of the blocks pulling it away from
the other. Find the position of the centre of mass at timet.
(A) Ft²/4m (B) Ft²/8m (C) Ft²/m (D) Ft²/2m​

Answer 1

The position of the centre of mass at time t is  x2 = 1/2 [(Ft)^2 / 2m−x0].

Explanation:

a) The acceleration of the centre of mass is

a.COM = F / 2m

The displacement of the centre of mass at time t will be

x = 1/2a.COM t^2 = Ft^2 / 4m

b) Suppose the displacement of the first block is x1 and that of the second is x2. Then,

x = mx1+mx2 / 2m  OR  Ft^2 / 4m = x1 + x2 / 2

x1 + x2 = Ft^2 / 2m ------- (1)

• Further, the extension of the spring is x1−x2. Therefore,

x1 − x2 = x0

From Eqs. (i) and (ii),

x1 = 1 / 2(Ft^2 / 2m + x0)

And x2 = 1/2 [(Ft)^2 / 2m−x0]

Thus the position of the centre of mass at time t is  x2 = 1/2 [(Ft)^2 / 2m−x0].

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