Initial Compressed Spring Force

In summary, a block of mass 0.058 kg is placed on a spring with a relaxed length of 0.17 m and a stiffness of 15 N/m. When the block is pushed down to a length of 0.11 m and held motionless, the force exerted by the spring on the block is 0.9 N in the +y direction. This can also be represented as a vector <0,0.9,0>N.
  • #1
Cpt. DeMorgan
4
0

Homework Statement



A spring has a relaxed length of 0.17 m, and its stiffness is 15 N/m. The spring sits vertically on a table. You place a block of mass 0.058 kg on the spring and push down on the block until the spring is only 0.11 m long. You hold the block motionless on the compressed spring, then remove your hand.

At the instant you release the block, what is the force exerted by the spring on the block? (Consider the initial stretch of the spring.) Use the usual axis system, with the +y axis running vertically up.

L=0.11m
L0=0.17m
Ks=15N/m

Homework Equations



|s|(absolute value of s)=|L-L0|

|Fspring|(magnitude of the force) = Ks*|s|

The Attempt at a Solution



I am asked to find the force exerted by the spring initially. I plugged in the variables I was given.

|s| = |0.11m-0.17m|
|s| = 0.06m

|Fspring| = 15N/m*0.06m
|Fspring| = 0.9N

So my answer for the force acted on the block initially by the spring was 0.9N. However, I am told this is not right. Can anyone explain why not?
 
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  • #2
Your method looks right. Are you being penalized for not enough SDs? You can get an extra SD by multiplying before subtracting.

[great user name, btw :) ]
 
  • #3
lewando said:
Your method looks right. Are you being penalized for not enough SDs? You can get an extra SD by multiplying before subtracting.

[great user name, btw :) ]
I think not.

15(0.17-0.11) = 15(0.17) - 15(0.11)

             = 2.55 - 1.65

Each of which is accurate to the nearest tenth (according to SigDig). So while each of these has 2 SigDig, when you subtract you're left with 1 SigDig.

However, if you multiply both lengths, 0.17 & 0.11, by 2.0000, before subtracting, the result will have 2 SigDig. Then if you multiply by 15 followed by division by 2.0000, the result will have 2 SigDig, showing the folly of being too rigid about applying the rules of SigDig.

Alternatively, if you look at the question, there does appear to be an emphasis on directionality. Should the answer of a magnitude be accompanied by a direction?
 
  • #4
Are you being penalized for not enough SDs?

I don't think my professor is that strict with SDs. It is possible though.

Should the answer of a magnitude be accompanied by a direction?

This is possible but isn't it implied to be going up the y-axis since it is a positive number? Maybe I should have wrote it as a vector: <0,0.9,0>N?
 
  • #5


Your solution is correct. The initial force exerted by the spring on the block is indeed 0.9N. It is possible that you made a calculation error or there was a typo in the problem. It is always a good idea to double check your calculations and make sure all the units are consistent. In this case, the units of length (m) and stiffness (N/m) should cancel out to give the correct unit of force (N). If you are still unsure, you can ask your teacher or a classmate for clarification.
 

1. What is Initial Compressed Spring Force?

Initial Compressed Spring Force, also known as initial compression force or initial force, is the force applied to a spring when it is compressed from its natural length or position.

2. How is Initial Compressed Spring Force calculated?

Initial Compressed Spring Force is calculated by multiplying the spring constant (k) by the amount of compression (x). The formula is F = kx, where F is the force in newtons, k is the spring constant in newtons/meter, and x is the compression in meters.

3. What factors affect the Initial Compressed Spring Force?

The Initial Compressed Spring Force is affected by the material and thickness of the spring, the degree of compression, and the surrounding temperature and pressure.

4. What is the unit of measurement for Initial Compressed Spring Force?

The unit of measurement for Initial Compressed Spring Force is newtons, which is a unit of force in the International System of Units (SI).

5. Why is Initial Compressed Spring Force important in scientific research and applications?

Initial Compressed Spring Force is important in scientific research and applications because it is a fundamental concept in the study of springs and their behavior. It is also crucial in the design and development of various devices and machines that utilize springs, such as suspension systems, shock absorbers, and mechanical watches.

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