Block dropped onto a spring

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A block of mass m = 1.8 kg is dropped from height h = 55 cm onto a spring of spring constant k = 1420 N/m (Fig. 8-36). Find the maximum distance the spring is compressed.

For this problem, I began by using the equation 2a(change in x) = (Velocity final) squared since velocity inital is simply 0. After that, I applied the equation 1/2m(v squared) to find the amount of kinetic energy, since looking at this system, there is no potential at the point where it hits the spring. Next, I said that the potential energy of the spring (1/2k (x squared)) + mg(change in y) = KE that I found from the first step. Next I formed a quadratic equation and found what the distance compressed was, but my answer is still incorrect, so I was just wondering what I was doing wrong, and if so, how to solve it.
 

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  • #2
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cd80187 said:
For this problem, I began by using the equation 2a(change in x) = (Velocity final) squared since velocity inital is simply 0. After that, I applied the equation 1/2m(v squared) to find the amount of kinetic energy, since looking at this system, there is no potential at the point where it hits the spring. Next, I said that the potential energy of the spring (1/2k (x squared)) + mg(change in y) = KE that I found from the first step. Next I formed a quadratic equation and found what the distance compressed was, but my answer is still incorrect, so I was just wondering what I was doing wrong, and if so, how to solve it.


You don't need to calculate kinetic energy. The initial energy is mgh. Assuming the spring to have negligible length, the finial energy is just 1/2(k)x^2. Equation these two will give you the spring displacement.
 
  • #3
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what about gravitational potential energy? and what would I use for the height for the initial energy?
 
  • #4
rsk
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Change in GPE of the block = EPE stored in the spring.

Just remember that the h you're using isn't just the 55cm. The block will fall further as the spring compresses (assuming that 55cm is the distance to the top of the spring?). SO h = 55cm + x.
 
  • #5
Andrew Mason
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cd80187 said:
Next, I said that the potential energy of the spring (1/2k (x squared)) + mg(change in y) = KE that I found from the first step. Next I formed a quadratic equation and found what the distance compressed was, but my answer is still incorrect...
Euclid, subject to the point made by rsk, is correct. The change in gravitational potential energy, mgh, is equal to the kinetic energy just before the spring starts compressing. That kinetic energy - plus a bit more gravitational potential as the spring compresses and the mass drops that compression distance -is then stored in the spring.

That gives you a quadratic equation that you should be able to readily solve.

AM
 
  • #6
BnJ
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So I am slow..... what equations do you use for this problem???
 

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