The spring in compressed until the initial sum of kinetic and potential energy equals the final stored energy in the spring, MINUS the decrease in PE during compression.
Let x be the maximum compression
(1/2) M Vo^2 + M g H = (1/2) kx^2 - M g x
Solve for maximum x deflection. When the spring is compressed x = 0.2 m, you can energy conservation to solve for v
(1/2) M Vo^2 + M g H = (1/2) kx^2 - M g x + (1/2) M v^2
A 5 kg mass is thrown down with a velocity of 10 m/s from a height of 20 m onto an upright spring with a relaxed length of 10 m, and a constant of 50 N/m. What will the maximum compression of the spring be? What will the speed of the object be when the spring is compressed 0.2 m?
Here's what I have so far:
INITIAL HEIGHT = 20 meters
FINAL HEIGHT = 10 meters
MASS = 5 kilograms
ACCELERATION = 9.8 meters per second squared
INITIAL VELOCITY = 10 meters per second
SPRING CONSTANT = 50 Newtons per meter
ENERGY OF GRAVITY = mgh
KINETIC ENERGY = 0.5mv^2
SPRING ENERGY = 0.5kx^2
TOTAL ENERGY = The sum of those 3 energies
I need to find the final velocity and then how much the spring gets compressed. Then I have to find the speed of the object when the compression of the spring is 0.2 meters.
1 answer
1 answer