A 0.280 kg block on a vertical spring with a spring constant of 4.12 × 10^
3 N/m is pushed downward, compressing the spring 0.0600 m.
When released, the block leaves the spring and travels upward vertically.
The acceleration of gravity is 9.81 m/s^
2. How high does it rise above the point of release?
Answer in units of m
2 answers
Just giving the equation to solve the problem would be nice too!
The potential energy of the compressed spring at release equals the potential energy gained at the maximum distance above the release point. Let that distnce be X and the initial spring compression be d = 0.06 m
(1/2) k d^2 = m g X
k is the spring constant, which you know is 4120 N/m.
m = 0.28 kg
Solve for X
(1/2) k d^2 = m g X
k is the spring constant, which you know is 4120 N/m.
m = 0.28 kg
Solve for X
2 answers