Spring Compression HW: Work, Height, Max Compression

[VitaX]

Homework Statement

A 506 g block is released from rest at height h0 above a vertical spring with spring constant k = 500 N/m and negligible mass. The block sticks to the spring and momentarily stops after compressing the spring 18.4 cm. How much work is done (a) by the block on the spring and (b) by the spring on the block? (c) What is the value of h0? (d) If the block were released from height 5h0 above the spring, what would be the maximum compression of the spring?

Homework Equations

Ws = -[.5kxf^2 - .5kxi^2]
Ug = mgh
K1 + Ug1 + Us1 = K2 + Ug2 + Us2

The Attempt at a Solution

a) Ws = -[.5*500*0^2 - .5*500*.184^2] = 8.464 J
b) Wb = -Ws = -8.464 J
c) Us = mgh; 8.464 = .506*9.8*h; h = 1.7069 m
d) mgh = .5kx^2 and solve for x using the value from c multiplied by 5 for h. I get .4114 m.

I'm not to sure about the equation for part c)
My reference height used was the spring compressed .184 m being 0. Though perhaps I should have set the initial height to zero (non compressed spring) and the final height to -.184 (compressed spring). What do you think?

 

[VitaX]

Ran my answer's through my online homework on WileyPLUS and parts a) and b) were correct. Parts c) and D) were incorrect which probably means part c) was incorrect.

Is part c) supposed to be setup as 8.464 = .506*9.8(h - .184) ?

 

[VitaX]

It seems the above wasn't it either as it says my answers for c) and d) are still incorrect. I only have 3 attempts to do each problem as well so I'm down to my last attempt. If anyone can help me correct my mistake with parts c) and d) that would be great.

 

[Doc Al]

Is part c) supposed to be setup as 8.464 = .506*9.8(h - .184) ?

Since h0 is the distance above the uncompressed spring, the total distance fallen by the mass = h0 + .184; measure the gravitational potential energy from the lowest point of the motion.

 

[VitaX]

Since h0 is the distance above the uncompressed spring, the total distance fallen by the mass = h0 + .184; measure the gravitational potential energy from the lowest point of the motion.

I see, I thought I had to find the distance above the noncompressed spring. Is my work or equation right for part d to solve for max compression at 5h0?

 

[Doc Al]

I see, I thought I had to find the distance above the noncompressed spring.

You do. That's h0. But the total height above the lowest point is h0 + .184.

Is my work or equation right for part d to solve for max compression at 5h0?

Why don't you rewrite the energy equation in terms of h0.

 

[VitaX]

You do. That's h0. But the total height above the lowest point is h0 + .184.


Why don't you rewrite the energy equation in terms of h0.

.5mv1^2 + mgh0 + .5kx^2 = .5mv2^2 + mgh0 + .5kx^2 (where h0 is 5 times h0 of part c)

it gets marked down to mgh0 = .5kx^2 which then becomes x = sqrt[(2mgh0)/k] then it's just plug and chug I believe. Hows that?

 

[Doc Al]

.5mv1^2 + mgh0 + .5kx^2 = .5mv2^2 + mgh0 + .5kx^2 (where h0 is 5 times h0 of part c)

it gets marked down to mgh0 = .5kx^2 which then becomes x = sqrt[(2mgh0)/k] then it's just plug and chug I believe. Hows that?

Seems like you're making the same error that you made with part c. So let's start there. Make sure that by h0 you mean the distance above the uncompressed spring. Write the energy equation for part c in terms of h0 and x, where x is the amount the spring gets compressed.

For part c, you're given x and solving for h0.

For part d, you'll replace h0 with 5h0 and then solve for x.

 

[VitaX]

Seems like you're making the same error that you made with part c. So let's start there. Make sure that by h0 you mean the distance above the uncompressed spring. Write the energy equation for part c in terms of h0 and x, where x is the amount the spring gets compressed.

For part c, you're given x and solving for h0.

For part d, you'll replace h0 with 5h0 and then solve for x.

Are you saying this for part C : .5mv1^2 + mg(h0 + .184) + .5kx^2 = .5mv2^2 + mgh + .5kx^2

0 + mg(h0 + .184) + 0 = 0 + 0 + .5kx^2

mg(h0 + .184) = .5kx^2 and solve for h0 for part c

Instead of doing as I had Ug = mg(h+.184) where Ug is my answer from part a

Part D :

.5mv1^2 + mg5h0 + .5kx^2 = .5mv2^2 + mgh + .5kx^2

0 + mg5h0 + 0 = 0 + 0 + .5kx^2

mg5h0 = .5kx^2 and solve for x (5h0 is my answer from part c multiplied by 5)

This is what I'm getting from your post above.

 

[Doc Al]

Are you saying this for part C : .5mv1^2 + mg(h0 + .184) + .5kx^2 = .5mv2^2 + mgh + .5kx^2

0 + mg(h0 + .184) + 0 = 0 + 0 + .5kx^2

mg(h0 + .184) = .5kx^2 and solve for h0 for part c

Instead of doing as I had Ug = mg(h+.184) where Ug is my answer from part a

Since Ug = .5kx^2, those results are equivalent.

I would write the general energy equation as:

mg(h0 + x) = .5kx^2

Part D :

.5mv1^2 + mg5h0 + .5kx^2 = .5mv2^2 + mgh + .5kx^2

0 + mg5h0 + 0 = 0 + 0 + .5kx^2

mg5h0 = .5kx^2 and solve for x (5h0 is my answer from part c multiplied by 5)

This is what I'm getting from your post above.

No, not correct. Look at my general equation for part c. The PE term must include all the height, not just the part above the uncompressed spring.

 

[VitaX]

Since Ug = .5kx^2, those results are equivalent.

I would write the general energy equation as:

mg(h0 + x) = .5kx^2


No, not correct. Look at my general equation for part c. The PE term must include all the height, not just the part above the uncompressed spring.

I'm sorry but I don't understand what you are saying is incorrect for part d. Because I thought it does include all the height since I wrote the correct equation for part c utilizing the spring compression as h0 + x. As for part D it's just the same exact equation from part c but the height is now increased 5 times. Could you write the equation for part d so I can understand what you are talking about.

 

[Doc Al]

I'm sorry but I don't understand what you are saying is incorrect for part d. Because I thought it does include all the height since I wrote the correct equation for part c utilizing the spring compression as h0 + x.

h0 + x is not the spring compression, x is. h0 is the height above the uncompressed spring; h0 + x is the height above the lowest point of the compressed spring.

As for part D it's just the same exact equation from part c but the height is now increased 5 times. Could you write the equation for part d so I can understand what you are talking about.

We must be talking past each other. It is the same equation:

For c: mg(h0 + x) = .5kx^2

For d: mg(5h0 + x) = .5kx^2

The only difference is that for d, h0 becomes 5h0.

 

[VitaX]

h0 + x is not the spring compression, x is. h0 is the height above the uncompressed spring; h0 + x is the height above the lowest point of the compressed spring.

We must be talking past each other. It is the same equation:

For c: mg(h0 + x) = .5kx^2

For d: mg(5h0 + x) = .5kx^2

The only difference is that for d, h0 becomes 5h0.

I see, so its quadratic:

mg5ho + mgx = .5kx^2 ------> .5kx^2 - mgx - mg5ho

x = (mg +/- sqrt[(mg)^2 - 4(.5k)(-mg5ho)])/2*.5k

Should I use the negative or positive answer as my final answer? I'm kind of thinking positive since they gave the compression in the question as positive.

 

[Doc Al]

I'm kind of thinking positive since they gave the compression in the question as positive.

Yes. Only the positive answer makes physical sense for this problem.