Question

In preparation for shooting a ball in a pinball machine, a spring k=700n/m, is compressed by .0742m relative to its unstrained length. The ball m=.0585kg, 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 .217m higher than point a. Ignore friction, and find the ball's speed at point b.

A) 1.23 m/s

B) 2.46 m/s

C) 3.69 m/s

D) 4.92 m/s

Answer 1

By using conservation of energy, the potential energy stored in the compressed spring is equated to the sum of gravitational potential energy and kinetic energy at point b. After calculation, the ball's speed at point b is found to be approximately 1.23 m/s, making option A the correct choice.

Step-by-step explanation:

To find the ball's speed at point b in a pinball machine, we need to use the principles of energy conservation. The potential energy stored in the compressed spring is converted into kinetic energy and gravitational potential energy as the ball rises to a higher point. When the spring is compressed by 0.0742 m and has a spring constant k=700 N/m, the potential energy stored in the spring is given by PEspring = 1/2 k x2, which can be converted into kinetic energy and gravitational potential energy as the ball moves from point a to point b.

The gravitational potential energy at point b given the vertical height difference is PEgravity = mgh, where m is the mass of the ball, g is the acceleration due to gravity, and h is the height difference. Using the conservation of energy, we can set the potential energy stored in the spring equal to the sum of the gravitational potential energy at point b and the kinetic energy of the ball at that point. Thus:

1/2 k x2 = 1/2 mv2 + mgh

Plugging in the given values:

1/2 (700 N/m) (0.0742 m)2 = 1/2 (0.0585 kg) v2 + (0.0585 kg) (9.8 m/s2) (0.217 m)

Solving for v, the ball's speed at point b, we find that v approximately equals 1.23 m/s. Therefore, option A) is the correct choice.