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Two blocks of masses 10 kg and 20 kg are connected by a massless spring and are placed on a smooth horizontal surface. A force of 200 N is applied on 20 kg mass as shown in the diagram. At the instant, the acceleration of 10 kg mass is 12 ms^-2, the acceleration of 20kg mass is,

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$4 {ms}^{- 2}$

$12 {ms}^{- 2}$

$20 {ms}^{- 2}$

$8 {ms}^{- 2}$

answer is A.

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Detailed Solution

For 10 kg block:

Tension in the spring: T = ma = 10(12) = 120 N

For 20 kg block:

$200 - T = ma \Rightarrow 20 (a) = 200 - 120 = 80 \Rightarrow a = 4 {ms}^{- 2}$

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