Question

A 2 kg block is connected with two springs of force constants k, = 100 N/m and
k = 300 N/m as shown in figure. The block is released from rest with the springs
unstretched. The acceleration of the block in its lowest position is (g= 10 m/s)
(1) zero
(2) 10 m/s- upwards
(3) 10 m/s2 downwards (4) 5 m/s2 upwards​

Answer 1

The acceleration of the block in its lowest position is a = 10 m/s^2

Explanation:

mgx - 1/2k1.x2 - 1/2k2.x2 = 0

20.x - 50.x^2 - 150 x^2 = 0

20 x = 200 x^2 x = 0.1 m

100 x 0.1 + 300 x 0.1 - 20 = ma

10 + 30 - 20 = 2a

40 - 20 = 2a

a = 20 / 2 = 10 m/s^2

Thus the acceleration of the block in its lowest position is a = 10 m/s^2

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