How Much Does the Spring Compress When a Man Jumps on a Scale?

In summary, the conversation discusses a man stepping on a scale and causing a spring inside to compress. The man then jumps and lands on the scale, causing further compression. The spring's maximum compression is 0.65 mm and the scale reading is equivalent to the man's weight. The conversation also mentions using the equations for spring constant and elastic potential to solve the problem.
  • #1
emilytm

Homework Statement


As a 75.0 kg man steps onto a scale, the spring inside the scale compresses by 0.65 mm. Excited to see that he has lost 2.5 kg since his last weighing, he jumps 0.3 m straight up into the air and lands directly on the scale.
What is the spring’s maximum compression?
What reading, in kilograms, does the scale give when the spring is at its maximum compression?

Homework Equations


Us=½k(Δl)2

The Attempt at a Solution


I used Us=½k(Δl)2 to find the spring constant, which I found to be 104.1N/m. Next, I found his total energy while he was jumping in the are, which was all in the form of gravitational potential energy. That was (9.8)(75kg)(.3m)=220.5J. That would also be equal to the kinetic energy of the man right before he makes contact with the scale after he jumps. I am not sure what to do next. Do I use the equation for elastic potential again to find the compression of the spring? I guess I don’t really know what maximum compression means. Thanks!
 
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  • #2
emilytm said:
I used Us=½k(Δl)2 to find the spring constant,
You did? But you you are not given initial energy, you are given a mass. You can do it with that equation but it seems a bit of a long way around. How about F=kx?
emilytm said:
104.1N/m.
kN/m?
emilytm said:
Do I use the equation for elastic potential
Yes, but be careful...
emilytm said:
equal to the kinetic energy of the man right before he makes contact with the scale after he jumps.
Ok, but bear in mind that gravity does not switch off at that point.
 

Related to How Much Does the Spring Compress When a Man Jumps on a Scale?

1. What is the "Spring Compression Problem"?

The "Spring Compression Problem" is a physics problem that involves calculating the displacement and force of a spring when a weight is placed on it. It is commonly used in introductory physics courses to demonstrate the relationship between force, displacement, and spring constant.

2. How do I solve the "Spring Compression Problem"?

To solve the "Spring Compression Problem", you will need to use Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement from its equilibrium position. You will also need to know the spring constant, which is a measure of the stiffness of the spring. By setting up and solving equations using these variables, you can determine the displacement and force of the spring.

3. What are the key components of the "Spring Compression Problem"?

The key components of the "Spring Compression Problem" are the spring itself, the weight or object placed on the spring, and the displacement and force of the spring. The spring constant is also an important component as it affects the amount of force exerted by the spring.

4. Can the "Spring Compression Problem" be applied to real-life situations?

Yes, the "Spring Compression Problem" can be applied to real-life situations. For example, it can be used to calculate the displacement and force of a car's suspension system when driving over a bumpy road. It can also be used to design and test springs for various industrial and mechanical applications.

5. What are some common misconceptions about the "Spring Compression Problem"?

One common misconception about the "Spring Compression Problem" is that the displacement and force of the spring will always be directly proportional. This is only true if the spring is within its elastic limit. Another misconception is that the weight on the spring will determine the displacement and force, when in fact, it is the spring constant that plays a larger role in these calculations.

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