An ideal Hookean spring of spring constant 20.0 N/m is connected to a 0.500 kg b

Question

An ideal Hookean spring of spring constant 20.0 N/m is connected to a 0.500 kg block in the arrangement shown to the right. The (*) represents the position of the center of the block when the spring is unstretched. (Positions are not drawn to scale.) From this position the experimenter slowly lowers the block from (*) until it reaches point B where the system remains at rest when the block is released. a. How much is spring stretched when the 0.500 kg block is at position B? b. The block is then pulled down to position A, 10.0 cm below position B. It is released and allowed to oscillate between positions A and C. In the space below, calculate the elastic energy, gravitational energy, and kinetic energy of the system at positions A, B, and C. Use position A as your zero height for the measuring of gravitational energy. Assume no dissipative forces. c. How fast is the block moving at the instant the center is even with position B?

Explanation / Answer

(a) Fnet = kx - m g = 0

20 (x) = 0.5 * 9.8

x = 0.245 cm .......Ans

(B) at A:

gravittaional PE = m g deltaH = 0

elastic PE = k h^2 /2 = 20 (0.245 + 0.10)^2 /2 = 1.19 J

KE = 0

at B:

Gravt. PE = 0.50 x 9.8 x 0.10 = 0.49 J

elastic PE = 20 (0.245^2)/2 = 0.6 J

KE = 1.19 - 0.49 - 0.6 = 0.10 J

at C:

GRavt PE = 0.50 x 9.8 x 0.20 = 0.98 J

elastic PE = 20 (0.245 - 0.10)^2 / 2 = 0.21 J

KE = 1.19 - 0.98 - 0.21 = 0 J

(C) KE = m v^2 /2

0.10 = 0.5 v^2 /2

v = 0.632 m/s

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