A 0.900 kg block is attached to a spring with spring constant 16.5 N/m . While t

Question

A 0.900 kg block is attached to a spring with spring constant 16.5 N/m . While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 32.0 cm/s. What is the blocks speed at the point where x=0.250A ? (find answer in cm/s)

Explanation / Answer

at the mean position x =0, hammer imparted kinetic energy 0.5 mv^2 = total energy = kinetic at x =0 TE = 0.5 *1*0.40*0.4 = 0.08 Joules when the spring stretches to the maximum, x = A total energy = elastic potential energy = 0.5 k A^2 --------------------------- energy conservation> 0.5 k A^2 = 0.08 A^2 = 0.08*2/k = 0.16/16 = 0.01 A = amplitude = 0.1 meter = 10 cm =================== let block speed be V, at x = A/2 = 0.05 m 0.5 kx^2 + 0.5 mV^2 = total energy = 0.08 V^2 = [0.16 - kx^2]/m V^2 = [0.16 - 16*0.05*0.0.05]/1 V^2 = 0.12 V = 0.346 m/s = 34.6 cm/s

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