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Question

A block of mass 4kg hangs from a spring of spring constant k = 400 N/m. The block is pulled down through 15 cm below and released. What is the kinetic energy when the block is 10 cm above the equilibrium position?


Your Answer
A
5J
Correct Answer
B
2.5J
Your Answer
C
1J
Your Answer
D
1.9J

Solution

The correct option is D 2.5J
Amplitude = 0.15 m
w=km=4004=10 rad/sf=w2π=5π Hz
Energy of the particle executing SHM = KE + PE
12mv2+12kx2=12KA2
Therefore, 12mv2=12kA212kx2=12(A2x2)
=12×400(0.1520.12)=2.5 J

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