Asked by HELP ME!

A pinball machine uses a spring that is compressed 4.0 cm to launch a ball. If the spring constant is 13 N/m, what is the force on the ball at the moment the spring is released?

I don't understand this I guess...
Does the ball leave the spring at the equalbarium position four cm from compression or does it travel past equalbarium and then the ball leaves? Also do I need to know the mass which I know how to do?

Could you please show me how to do this problem and the formulas to use... Thanks for the help
Answered by Damon
At the moment of release, the full force of the compressed spring is on the ball.
F = k x = 13 N/m * .04 m
Answered by cassie
-.52 N
Answered by Hi
F=kx=13N/m*.04m=.52 newtons
Answered by yareli
When an acrobat reaches the equilibrium position, the net force acting along the direction of motion is zero. Why does the acrobat swing past the equilibrium position?
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