Question

In preparation for shooting a ball in a pinball machine, a spring (k = 730 N/m) is compressed by 0.0796 m relative to its unstrained length. The ball (m = 0.0585 kg) is at rest against the spring at point A. When the spring is released, the ball slides (without rolling). It leaves the spring and arrives at point B, which is 0.256 m higher than point A. Ignore friction, and find the ball's speed at point B.

Answer 1

Answer:

The speed of the ball at point B is 8.59 m/s

Explanation:

This problem can be solved using the law of conservation of energy. The elastic potential energy stored in the ball at rest against the spring is converted into gravitational potential energy and kinetic energy. Since there are no dissipative forces (like friction), the sum of gravitational potential energy plus kinetic energy at point B must be equal to the elastic potential energy at point A.

The elastic potential energy can be calculated as follows:

EPE = 1/2 · k · x²

Where:

EPE = elastic potential energy.

k = spring constant.

x = displacement of the spring.

Then:

EPE = 1/2 · k · x²

EPE = 1/2 · 730 N/m · (0.0796 m)²

EPE = 2.31 J

At point B, the sum of gravitational (PE) and kinetic energy (KE) will be equal to the elastic potential energy at point A.

EPE = PE + KE

The gravitational potential energy is calculated as follows:

PE = m · g · h

Where:

m =  mass of the ball.

g = acceleration due to gravity.

h = height.

Then:

PE = m · g · h

PE = 0.0585 kg · 9.81 m/s² · 0.256 m

PE = 0.147 J

Now, we can calculate the kinetic energy:

EPE = PE + KE

EPE - PE = KE

2.31 J - 0.147 J = KE

KE = 2.16 J

Using the equation of kinetic energy, we can calculate the speed of the ball at point B:

KE = 1/2 · m · v²

Where:

m = mass of the ball.

v = velocity.

Then:

KE = 1/2 · m · v²

2.16 J = 1/2 · 0.0585 kg · v²

2 · 2.16 J / 0.0585 kg = v²

v = 8.59 m/s

The speed of the ball at point B is 8.59 m/s