Question

In the figure shown masses of the blocks A, B, C are 6 kg, 2 kg and 1 kg respectively. Mass of the spring is negligibly small and its stiffness is 1000 N/m. The coefficient of friction between the block A and the table is 0.8 Initially block C is held such that spring is in relaxed position. The block is released from rest. Find: (g=10m/s^2)

(a) the maximum distance moved by the block C

(b) the acceleration of each black, when elongation in the spring is maximum.

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Answer 1

Answer:

2×10 to the power -2 cm

Explanation:

Equilibrium position for block C is at a distance x=mg/k distance below the initial point. x=0.01m. If the block A doesn't move then the maximum distance C can move down is 0.02m under SHM. When C is moved to maximum distance then the tension in the spring is 1000 x 0.02N=20N.

If block A didn't move then block B also didn't move, thus

mg+kx

2

−T=0, where T is the tension in the string and x

2

is elongation in the spring.

T=mg+kx

2

=20+20=40N

For block A, maximum static friction is μm

A

g=48N, thus block

A won't move and maximum displacement of the block C is 20cm.

hope it helps