First slide

Second law

Question

Two masses 10 kg and 20 kg respectively are connected by a massless spring as shown in figure. A force of 200 N acts on the 20 kg mass. At the instant shown in figure, the 10 kg mass has acceleration of 12 m/s². The value of acceleration of 20 kg mass is

Easy

A

4 m/s²
B

10 m/s²
C

20 m/s²
D

30 m/s²

Solution

$Given, m_{1} = 10 kg, m_{2} = 20 kg, F_{2} = 200 N, a_{1} = 12 m / s^{2}, a_{2} = ?$
For mass $m_{1}, F_{1} = m_{1} a_{1}$ = 10 x 12 = 120 N
For mass m₂,
$F_{2} - F_{1} = m_{2} a_{2}$
$200 - 120 = 20 \times a_{2}$
$\Rightarrow a_{2} = \frac{80}{20} = 4 {ms}^{- 2}$

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