Question

Given the following about a spring:

mass = 20.0 kg
height initial = 5.25 m (above the ground)
height final = 2.17 m (above the ground with the mass attached)
delta y initial = 0.00 m

There are several questions I must answer from this.

1. What is the value of the spring constant?

I have solved this with the formula kx = mg, which gave me the value of 63.7 N/m as k.

2. What would the velocity of the falling mass be half way down?

3. How much work would it take to return the mass to its original position?

Answer 1

\[v ^{2}=u ^{2}+2as\] where v is the final velocity,
u is the initial velocity,
a is accelaration, which in this case would be g,
and s is the distance, which in this case would be halfway down.

Answer 2

Work done is equal to the -(Change in Energy).
So calculte the porential energy, at the lower stage, and calculate the potential energy energy at the initial stage, and subtract them.
Since initail stage has greater energy, you shouuld get a ositive number.
Do tell if you are still in doubt.

Answer 3

Therefore, wouldn't the initial velocity be zero because the object started from rest? Also, would halfway down be equal to 5.25-2.17 / 2? Thanks

Answer 4

Yes and Yes.
Very good.

Answer 5

Doing the following gives me the equation \[v =\sqrt{2*9.81*1.54} = 5.5 m/s\]

Answer 6

Now on to the second question

Answer 7

So, the potential energy for when the spring is fully stretched is found with the formula
\[PE = 1/2kx^2\]
\[PE = 1/2(63.7)(3.08)^2\]
PE = 302.14184 J

Answer 8

That's for when stretched. The 3.08 comes from the distance of 5.25 - 2.17. Is that correct?

Answer 9

Yes, looks correct.

Answer 10

So now, to get the PE at the initial stage, I add the 2.17 to 5.25 to get the total distance up and then substitute that number into the equation?

Answer 11

Then of course I would subtract the final from the initial

Answer 12

Yes.

Answer 13

Hold on.... I might have made a mistake with velocity question....

Answer 14

Sure, I've got time. Thanks sooo much for helping me!

Answer 15

The formula is for a free falling body.

Try this, in a spring, the potential energy is equal to the kinetic energy half way.
So calculate the potential energy at half way, and equate it to
1/2kx^2, which is also equal to 1/2mv^2.

from this you can take out the velocity.

Answer 16

Cool, np.

Answer 17

So \[1/2(63.7)(1.54) = 1/2(20.0)v^2\]

Answer 18

Would I still use the 1.54 as half of the distance when stretched? (3.08)

Answer 19

That gives me 2.75 m/s as the velocity

Answer 20

Finally, for question 3:
PE = 1/2(63.7)(7.42^2)
PE = 1/2(63.7)(55.0564)
PE = 1753.55 J

Thus, subtracting 302 from this gives me: 1452 J required to move it back to its original position.

Answer 21

7.42?

Answer 22

5.25 + 2.17 to be the total distance above the ground. The number seems off to me though.

Answer 23

Oh wait, nvm. It's supposed to be 5.25 because the 2.17 is when stretched.

Answer 24

that's not the total distance....
this is a spring.
the initial height, is the max height.
the second height is just by coming closer to the ground
imagine a fixed slinky, suspended in air

Answer 25

Ah, my bad. So the initial height of 5.25 is what I should have used. When I do the equation with that number, I get: 877.87J

Answer 26

Then, subtracting these two number, he result is: 576 J required to move the object back.

Answer 27

sounds correct.
My advice would be to verify these results, but I think they are correct.

Answer 28

Thank you so much. If you have time, I do have one more question. How much power would it take to return the mass to its original position in 3.5 seconds?

Answer 29

Power is work done divided by me \[p=w/t\]
you have time, and work done.

Answer 30

You're the best!

Answer 31

:-)