What is the speed of the block when the spring is compressed by 10 cm?

In summary: Kf + Us = Ki + Ui1/2mv^2 + 1/2kx^2 = mgh (Ki = 0 so it is dropped)[move everything over so you get v alone]v = [sqrt] (2gh - (kx^2/m))v = [sqrt] (2*9.81*8 - (1086*.1^2/9.3))v = 12.56 m/s I get your answer using 8.1 m, the complete descent of the block to this point. (If you use only 8.0 m, you get v = 12.48 m/sec, which is probably still within the
  • #1
grouchy
73
0
A 9.3 kg block is dropped onto a spring of spring constant 1086 N/m from a height of 800 cm. When the block is momentarily at rest, the spring has been compressed by 50 cm. Find the speed of the block when the compression of the spring is 10 cm. The acceleration of gravity is 9.81 m/s^2. Answer in units of m/s.


My attempt:

Ki + Ui = Kf + Us
Ki = o so...
Kf = Ui - Us
1/2 mv^2 = mgh - 1/2kx^2

v = (sqrt)mgh - 1/2kx^2/2m
v = (sqrt)(9.3)(9.81)(8) - 1/2(1086)(.1)^2/2(9.3)
v= 6.240838447m/s which is wrong.
 
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  • #2
grouchy said:
A 9.3 kg block is dropped onto a spring of spring constant 1086 N/m from a height of 800 cm. When the block is momentarily at rest, the spring has been compressed by 50 cm. Find the speed of the block when the compression of the spring is 10 cm. The acceleration of gravity is 9.81 m/s^2. Answer in units of m/s.


My attempt:

Ki + Ui = Kf + Us
Ki = o so...
Kf = Ui - Us
1/2 mv^2 = mgh - 1/2kx^2

v = (sqrt)mgh - 1/2kx^2/2m
v = (sqrt)(9.3)(9.81)(8) - 1/2(1086)(.1)^2/2(9.3)
v= 6.240838447m/s which is wrong.

This is a favorite homework or exam problem because it catches just about everybody out. Don't forget that the block continues to descend as the spring is compressed, so the change in gravitational potential energy is a bit larger than what you get from falling just 8.00 m.

Be careful also about how you divided out that 1/2 when you were solving for v. There is at least one algebra error...
 
  • #3
uhh... so should it be 8.5m instead of just 8? and with the 1/2 fix it should be

v = (sqrt) 2(9.3)(9.81)(8.5) - (1086)(.1)^2/(9.3) ?(on my last try so i can't afford to screw it up again :(
 
Last edited:
  • #4
grouchy said:
uhh... so should it be 8.5m instead of just 8? and with the 1/2 fix it should be

v = (sqrt) 2(9.3)(9.81)(8.5) - (1086)(.1)^2/(9.3) ?(on my last try so i can't afford to screw it up again :(

I take it you're on one of those computer-entry problem systems. Could you check the problem statement again, please? As I've been working through the numbers, the spring constant seems inconsistent with the other information in the problem. (This has become important now that you tell me you're down to one more try.)
 
  • #5
m = 9.3 kg
Dropped from 800 cm = 8 m
gravity = 9.81 m/s^2
Spring Constant = 1086 N/m
When the block is momentarily at rest, the spring has been compressed by 50 cm = .5 m

and the question asks for the speed of the block when the compression of the spring is 10 cm = .1 m

those are all the values given in the problem
 
  • #6
grouchy said:
m = 9.3 kg
Dropped from 800 cm = 8 m
gravity = 9.81 m/s^2
Spring Constant = 1086 N/m
When the block is momentarily at rest, the spring has been compressed by 50 cm = .5 m

and the question asks for the speed of the block when the compression of the spring is 10 cm = .1 m

those are all the values given in the problem

All right, here's the thing. The energy you get by dropping the block by 8 meters should compress the spring a lot more than 0.5 m. I can find no way to make the first part of the problem consistent with the second part by adjusting any of the numbers in a way that could be caused by a typographical error. You may need to talk to the course instructor about this one. (It wouldn't be the first time there was a mistake in a computer system problem...)

In fact, the problem is really overdetermined as it stands. You don't need more than two of the following: the height from which the block is dropped, the spring constant, or the maximum compression distance of the spring. One of the three values for these in the problem would seem to be in error.

Compare the potential energy released by the block falling 8 + 0.5 meters with the spring potential energy when it is compressed 0.5 meters. There is a substantial discrepancy -- or are we supposed to consider that there is a large transformation of mechanical energy?

If you must solve this problem during the weekend, I'd take a chance that the maximum compression distance and spring constant are correct and use that information to find the speed of the block when the spring was compressed by 0.1 m.

I'm sorry I can't be more helpful than this, but it looks like there is a mistake in the statement of the problem. (That wouldn't be the only time that's happened this month...)
 
  • #7
Hey, thanks for your help. I managed to get the correct answers too lol. Here it is for someone who might have a similar problem.

Kf + Us = Ki + Ui
1/2mv^2 + 1/2kx^2 = mgh (Ki = 0 so it is dropped)

[move everything over so you get v alone]

v = [sqrt] (2gh - (kx^2/m))
v = [sqrt] (2*9.81*8 - (1086*.1^2/9.3))
v = 12.56 m/s
 
  • #8
grouchy said:
v = [sqrt] (2gh - (kx^2/m))
v = [sqrt] (2*9.81*8 - (1086*.1^2/9.3))
v = 12.56 m/s

I get your answer using 8.1 m, the complete descent of the block to this point. (If you use only 8.0 m, you get v = 12.48 m/sec, which is probably still within the accepted precision for the answer.)

If you use this to find the compression at which the velocity will be zero, you get about 1.25 m (which is about what I was finding earlier), with the block descending a total of 9.25 m. So the statement about where the block is momentarily at rest should be discarded from the problem as it is, because the 50 cm. figure is wrong (and unneeded)...
 
  • #9
I used this equation for my exact same problem and got it wrong twice so far. I only have two tries left.

A 2.45 kg block is dropped onto a spring with a force constant of 4223 N/m from a height of 4.82 m. When the block is momentarily at rest, the spring is compressed by 24.0 cm. Find the speed of the block when the compression of the spring is 13.3 cm.

I even reworked the problem myself and it makes total sense to me. I ended up with 9.71m/s as my answer yet it says it's incorrect.

Am I missing something here?
 
  • #10
ok figured it out. You do have to add the amount of spring compression to the height.
 

1. How does a block falling on a spring affect the spring's motion?

When a block falls on a spring, it compresses the spring due to the force of gravity acting on the block. This compression of the spring causes it to store potential energy.

2. What is the relationship between the mass of the block and the amount of compression of the spring?

The amount of compression of the spring is directly proportional to the mass of the block. This means that the heavier the block, the more it will compress the spring.

3. What happens to the potential energy stored in the spring when the block is released?

When the block is released, the potential energy stored in the spring is converted into kinetic energy as the spring expands back to its original length. This kinetic energy then causes the block to bounce off the spring.

4. How does the height at which the block is dropped affect the spring's motion?

The height at which the block is dropped affects the potential energy of the block-spring system. The higher the block is dropped from, the more potential energy it will have. This results in a greater compression of the spring and a higher bounce for the block.

5. How does the stiffness of the spring affect the block's motion?

The stiffness of the spring, also known as its spring constant, determines how much force is required to compress or stretch the spring. A stiffer spring will result in a smaller amount of compression when the block falls on it, while a less stiff spring will compress more. This affects the height of the block's bounce off the spring.

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