Student pushes a baseball 0.16kg down onto top of a vertical spring that has its lower end fixed to a table, compressing the spring a distance of 0.25m. The spring constant of the spring is 135N/m.

Question

a) The ball is then released. What is its speed, v, in meters per second, just after the ball leaves the spring?

b) What is the maximum height, h, in meters, that the ball reaches above the top of the uncompressed spring?

Answer 1

When the spring is compressed, it stores potential energy given by PE = ½ k X^2 where k is the spring constant , and X is the amount compressed.

PE = ½ x 135 N/m x 0.25m^2 = 4.22 Joules.

This PE is converted to Kinetic Energy which is equal to ½ M V^2.

Equate the PE to KE and solve for V.

V = sq root ( 2 x PE/M) = 7.26 m/s

The height to which the ball reaches can be obtained from the equation Vf^2 = Vo^2 – 2gH, where Vo and Vf are the initial and final velocities respectively, g is 9.8 m/s^2, and H is the height attained.

Solve for H, noting that the final velocity is zero.

H = 0 –(7.26^2) / (-2 x 9.8 x 7.26) = 0.37 meters