Question

A 0.850 kg block is attached to a spring with spring constant 18.0 N/m . While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 31.0 cm/s . What are

The amplitude of the subsequent oscillations?

The block's speed at the point where x= 0.150 A?

Answer 1

at the mean position x =0, hammer imparted kinetic energy

0.5 mv^2 = total energy = kinetic at x =0
TE = 0.5 *1*0.85*0.31*0.31 = 0.0408 Joules
when the spring stretches to the maximum, x = A
total energy = elastic potential energy = 0.5 k A^2
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energy conservation>
0.5 k A^2 = 0.0408
A^2 = 0.0408*2/k = 0.0816/18 = 0.0673
A = amplitude = 6.7 cm
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let block speed be V, at x = 0.15A = 0.15*0.0673 = 0.0100
0.5 kx^2 + 0.5 mV^2 = total energy = 0.0408
V = 0.3052m/s

V= 30.52cm/s