Question

A 2 meter long mass less spring is hanging from the ceiling. You hang a 1 kg weight to the bottom of the spring. After the weight is attached the spring stretches 5 cm down. You take off the weight and removed the spring from the ceiling and place it on the floor so it is horizontal.

1. There is no friction on the floor. Calculate the maximum compression that the spring reaches at its relaxed position if you fire a 0.5 kg weight at it with a velocity of 20 meters per second.

2. How much would the answer to the first question change if the spring was only stretched 1 cm downward when the 1 kg weight is attached?

3. If the spring from question 2 is compress a distance equal to spring from question 1 and then put the 0.5 kg weight there and released it, what would be the velocity of the mass after it left the spring?

Answer #1

Vertical force balance gives, kx = mg

=> k = mg/x_{o} = 1 * 9.8 / 0.05 = 196 N/m

1) KE of the 0.5 kg weight is converted to the elastic PE of the spring.

mv^{2}/2 = kx_{max}^{2}/2

=> x_{max} = v(m/k)^{1/2} = 20 * (0.5 /
196)^{1/2} = 1.01 m

2) k_{new} = 1 * 9.8 / 0.01 = 980 N/m

x_{max}' = v(m/k_{new})^{1/2} = 20 *
(0.5 / 980)^{1/2} = 0.45 m

3) Elastic PE of the spring is converted to the KE of the 00,5 kg weight.

mv^{2}/2 =
k_{new}x_{max}^{2}/2

=> v = x_{max} (k_{new}/m)^{1/2} =
1.01 * (980 / 0.5)^{1/2} = 44.7 m/s

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