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?

Question

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

Answer 1

At the moment of release, the full force of the compressed spring is on the ball.
F = k x = 13 N/m * .04 m

Answer 2

-.52 N

Answer 3

F=kx=13N/m*.04m=.52 newtons

Answer 4

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?