How Do You Calculate the Compression of a Spring When a Block is Dropped on It?

[jamesm113]

Hello,
This is my first post, so please forgive any mistakes I might have made. I am having a little difficulty with these 3 problems. Any push would be much apperciated. Thanks!

1. A 2.0 kg block is dropped from a height of 80 cm onto a spring of force constant k = 1960 N/m, as shown in Fig. 12-19. Find the maximum distance the spring will be compressed.
I figured that energy due to gravity must equal the energy of the spring - in other words, ky^2/2 = mgy.

2.Tarzan, who weighs 655 N, swings from a cliff at the end of a convenient vine that is 21 m long (Fig. 12-21). From the top of the cliff to the bottom of the swing, he descends by 3.2 m. The vine will break if the force on it exceeds 950 N. What is the greatest force on the vine during the swing?
I figure that greatest force will be when the force of gravity adds up with the force of the circuluar motion. However, I do not believe there is enough information to compute the velocity of the swing.

 

[Hootenanny]

Hi there James and welcome to PF,

You're answer to question one is spot on, just a minor point however you stated the equation as ¹/₂ky² = mgy; this is not correct because the compression of the spring will not equal the height from which the block was dropped, you're equation should be ¹/₂kx² = mgy where x is the compression. As for question two, there most certainly is enough information to calculate the velocity at the lowest point, which is where the tension will be greatest. Think about potential energy.

~H

 

[al_201314]

Just wondering, why would it be kx^2 since the work done if the spring is stretch or compressed is = 1/2(k)(x)^2 which is the gain in kinetic energy of the block from the drop?

 

[Hootenanny]

Just wondering, why would it be kx^2 since the work done if the spring is stretch or compressed is = 1/2(k)(x)^2 which is the gain in kinetic energy of the block from the drop?

Ahh, sorry thank you for the correction al. I just missed the half off by accident . I have duely recitified my error.

~H

 

[jamesm113]

On the first one, I've tried solving for the equation 1/2ky2 = mgy, which worked out to be x = sqrt(2mgy/k). Plugged in all the numbers sqrt((2*2*9.8*.8)/1960), and I got .12649, which did not work.

 

[Hootenanny]

On the first one, I've tried solving for the equation 1/2ky2 = mgy, which worked out to be x = sqrt(2mgy/k). Plugged in all the numbers sqrt((2*2*9.8*.8)/1960), and I got .12649, which did not work.

I apologise, I forgot to add on the additional potential energy loss due to the spring compressing. The formula should be;

\frac{1}{2}kx^2 = mg(h+x)

The block will lose additional potential energy when the spring is compressed. If you expand the above expression out, you should obtain a quadratic equation in terms of x.

Apologies for the mix up. Do you undersand where the formula come from? The (h+x) is the height from which the ball was dropped plus compression of the string, which represents the total loss of potential energy.

~H