A physics question about a spring?

Question

A spring is used to launch a ball with a mass of 1.00 x 10^-2 kg. The spring is compressed 10.0cm using an average force of 35.0 N. For a launch, the ball is placed against the movable end of the spring and the compressed spring is released.

Suppose the spring launches the bal sideways along a track. When the spring is realeased, what is the kinetic energy of the ball the moment it leaves the end of the spring?

A. 1.23 J

B. 1.75 J

C. 2.29 J

D. 3.50 J

What is the speed of the ball the moment it leaves the end of the spring?

A. 13.3m/s

B. 26.5m/s

C. 85.7m/s

D.700m/s

Suppose the ball is launched straight up into the air. How high will the ball go?

A. 0.100m

B. 10.1m

C. 17.8m

D.35.7m

Thanks answer what you know, THANKS!

Answer 1

Using:

k = spring constant

x = distance of compression

The average force needed to compress a spring is:

F = (1/2)kx

35 = (1/2)k(0.1)

k = 700 N/m

Energy stored in a compressed spring is just:

E = (1/2)kx^2

E = (1/2)(700)(0.1)^2

E = 3.5 J

And this equals the kinetic energy when the ball is launched. So the answer to the first question is D.

The speed of the ball can be found using the relation for kinetic energy.

E = (1/2)mv^2

3.5 = (1/2)(0.01)v^2

v^2 = 700

v = 26.45 m/s

So the speed of the ball is about 26.5 m/s and the answer is B.

First find the time it takes the ball to reach max height:

v = v0 + at

At max height v = 0 so:

at = v0 and t = v0/a = 26.5/9.8 = 2.7 seconds

Distance travelled is given by (g is negative since it is acting in the opposite direction to the ball):

s = v0t + (1/2)gt^2

s = (26.45)(2.7) - (1/2)(9.8)(2.7)^2

s = 71.415 - 35.721

s = 35.7 meters

So the answer to the last question is D.