Asked by Liz
A 8.1 kg stone is at rest on a spring. The spring is compressed 12 cm by the stone. The stone is then pushed down an additional 27 cm and released. To what maximum height (in cm) does the stone rise from that position?
Answers
Answered by
drwls
8.1 kg weighs M g = 73.4 N; so the spring constant must be k = 73.4/0.12 = 661.5 N/m
After compressing another X = 0.27 m, the releasable potential energy in the spring is (1/2)kX^2 = 24.1 J
Set 24.1 J = M g H to get the height H that the stone can rise.
H = 24.1 J/(8.1*9.8) = 0.30 m
After compressing another X = 0.27 m, the releasable potential energy in the spring is (1/2)kX^2 = 24.1 J
Set 24.1 J = M g H to get the height H that the stone can rise.
H = 24.1 J/(8.1*9.8) = 0.30 m
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