What is the Coefficient of Friction in Conservation of Energy Problem?

In summary, a spring with a constant k=10[N/m] is attached to a weight of 0.05 kg on a table. The string is stretched 10 cm and released, causing the weight to travel 40 cm until it halts. Using the elastic energy equation and conservation of energy, the coefficient of friction is calculated to be μ=0.246. However, after considering the installation of the system and the possibility of a different distance traveled by the weight, it is determined that the correct answer is μ=0.2.
  • #1
Karol
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Homework Statement


A spring with a constant k=10[N/m] is attached with a long rope to a weight of 0.05 kg. they both lie horizontally on the table. the string is stretched 10 cm and released and the weight travels 40 cm on the table until it halts. what is the coefficient of friction

Homework Equations


Elastic energy of spring:[itex]E_P=\frac{1}{2}kx^2[/itex]

The Attempt at a Solution


Conservation of energy:
[tex]\frac{1}{2}kx^2=mg\mu\cdot \triangle x[/tex]
[tex]\frac{1}{2}10\cdot 0.1^2=0.05 \cdot 10 \cdot \mu \cdot 0.4 \rightarrow \mu=0.25[/tex]
The answer should be μ=0.2
 

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  • #2
With friction present, the spring may end up still stretched a bit.
 
  • #3
I don't see how this can happen in this case since the spring gives velocity to the mass which continues travelling.
In any case the initial potential energy of the spring is the same, it is stretched 10 cm.
 
  • #4
If the spring remains stretched a bit, it retains some potential energy.
 
  • #5
The spring will stop at a distance of:
[tex]F=mg\mu=kx\rightarrow 0.5\mu=10x \rightarrow x=0.05\mu[/tex]
Conservation of energy:
[tex]\frac{1}{2}k \left(0.1^2-0.05^2\mu^2\right)=mg\mu\triangle x[/tex]
[tex]5(0.01-0.0025\mu^2)=0.5\mu 0.4[/tex]
[tex]0.0125\mu^2+0.2\mu-0.05=0 \rightarrow \mu=0.246[/tex]
It still isn't μ=0.2
 
  • #6
Right now I don't see how the problem can be solved unless the coefficient of static friction is also given.

We start to pull on the spring. At first, the block stays put. At some spring stretch level the block starts to move. This will happen long before the full 10 cm of spring tension is reached since at 10cm the spring force is 1N, way more than what is required to start the weight moving even for a high static friction coefficient. So we have to continue pulling on the spring even when the weight has started to accelerate, until the full 10cm of spring tension is reached. So the puller applies not only the energy in a 10cm-expanded spring but also his force x distance until the 10 cm spring extension is reached.

That is the total system input power. The system output power is of course weight x kinetic friction coefficient x distance traveled after the spring is released.

A pretty complicated problem and I wonder if the wording is correct.
 
  • #7
rude man said:
We start to pull on the spring. At first, the block stays put. At some spring stretch level the block starts to move.
The system isn't like that. you don't start to pull the spring until the mass also moves, no. you pull the mass which pulls the string through the rope. i knew my simple drawing isn't good enough.
You pull the mass, which stretches the spring, a distance of 10cm and release.
 
  • #8
Installation

There can be no residual elongation of the spring because of the installation. the spring is pulled by a long cord. at the end of the cord is the mass. the mass is released and it travels until it stops, in front of the spring.
 

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  • #9
Karol said:
There can be no residual elongation of the spring because of the installation. the spring is pulled by a long cord. at the end of the cord is the mass. the mass is released and it travels until it stops, in front of the spring.
Right. I hadn't looked at this closely enough. It' doesn't seem complicated. The spring gives up all its energy. The mass travels 40cm.

So your calculations seem right. μk = 0.255

Maybe one of the data values isn't right? e.g., say the mass travels 51cm ...
 
  • #10
OK, with this picture I don't see why your original answer isn't correct. Hope you can post what you found out later about it.

EDIT: I'll bet the spring traveled 40 cm from its original position. That makes Δx = 0.4 + 0.1 = 0.5 and then the answer is indeed μ = 0.2.
 
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1. What is the law of conservation of energy?

The law of conservation of energy states that energy cannot be created or destroyed, it can only be transformed from one form to another. This means that the total amount of energy in a closed system remains constant over time.

2. Why is the conservation of energy important?

The conservation of energy is important because it is a fundamental principle in physics and helps us understand the behavior and interactions of various systems. It also allows us to make accurate predictions and calculations in many scientific fields.

3. What are some examples of energy conservation?

Some examples of energy conservation include turning off lights when leaving a room, using energy-efficient appliances, and carpooling to reduce fuel consumption. Another example is the use of renewable energy sources, such as solar and wind power, to reduce our reliance on fossil fuels.

4. Is energy conservation the same as energy efficiency?

No, energy conservation and energy efficiency are not the same. Energy conservation refers to reducing energy consumption, while energy efficiency refers to using energy in a more efficient way to achieve the same results. For example, using LED light bulbs instead of incandescent bulbs is an energy efficiency measure, while turning off lights when not in use is an energy conservation practice.

5. How can we apply the law of conservation of energy in our daily lives?

We can apply the law of conservation of energy in our daily lives by being mindful of our energy consumption and taking steps to reduce it. This can include using public transportation or carpooling instead of driving, turning off lights and unplugging electronics when not in use, and using energy-efficient appliances. We can also support renewable energy sources and advocate for sustainable practices in our communities.

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