Question

A vertical spring(ignore its mass), whose spring constant is 875N/m, is attached to a table and is compressed down by .160m. What upward speed can it give to a .350kg ball when released?How high up its original position (spring compressed) will the ball fly?

Answer 1

I'm still a bit confused about setting this problem up...

Answer 2

I know when its compressed it has 0 velocity and gravity is acting on the spring by the ball going downward, and when its released, the spring is stretched upwards

Answer 3

I know I can't use this formula for elastic potential energy:
\[\frac{1}{2}m v^{2}+\frac{1}{2}kx^2=\frac{1}{2}mv^2+\frac{1}{2}k^2\]
since gravity is accounted for in the first point. Would I have to combine the above formula with the mechanical energy one:
\[\frac{1}{2}mv^2=\frac{1}{2}mv^2+mgy\]
to get something like this:
\[\frac{1}{2}m v^{2}+\frac{1}{2}mgy=\frac{1}{2}mv^2+\frac{1}{2}k^2\]?

Answer 4

Since the velocity at this point, when the spring is compressed, is assumed to by 0, the 1/2mv^2 wouid cancel out. Since the ball is compressed down, I'm guesing that .160m would be negative.

Answer 5

delta s is the spring right?

Answer 6

ok

Answer 7

Choose as origin of altitudes the initial position of the mass (when the spring is compressed).
It is easier to answer the second question first (independantly of the first one) because elastic potential energy is simply converted into gravitational potential energy (velocities are 0 in both situations).

Answer 8

Okay. I'll try this later on when I have time and get back to you guys with any issues I may have. Thanks.

Answer 9

okay so I ended up getting 7.80m/s

Answer 10

Correct ^_^

Answer 11

That second equation says:

\[y _{1}=\frac{v _{0}^2}{2g}\]
right

Answer 12

where y1 is the height?

Answer 13

You should also add the 16 cm , as the question reads : "How high up its ORIGINAL POSITION (spring compressed) will the ball fly?"

You can also solve this problem comparing initial and final states and not bothering about intermediate speed.

Answer 14

If xₒ is the initial compression and h is the height reached above this point, then conservation of energy implies that:\[\frac{1}{2}kx _{o}^{2} = mgh\]Working out h is dead-easy!

Try it and you will have the same value as your y1 + 0.16

I have to go now, but what you did is fine and you do not need help any more.

Answer 15

Okay thanks, I ended up getting 3.26m your method, and 2.85 the other way.

Answer 16

Your method gives y1 = 7.8²/(2x9.8) = 3.10 m
Add 0.16 m and you have h = 3.26 m as well !

Answer 17

I see now, I plugged in the wrong numbers.