A spring is shortened to 2 cm by a force of 0.6N
a- Calculate the spring constant of the spring.
b- What length variation produces a piece of iron of 200 \(cm^3\) volume?
c- What is in this case its final length (p iron = 7800 kg/ \(m^3\)?

A spring has a length of 40 cm at rest, we hang to its free extremity an empty bottle, its length becomes 42 cm, we fill the bottle by 1.5 liter, the final length of the spring is 50 cm.
What is the weight of the empty bottle? (p pure alcohol = 800 kg/ m^3. Please explain your solution in details.

A spring has a length of 30 cm at rest and it elongates by 2cm for a weight of 20N. The spring retains its perfect elasticity as its length does not exceed 40 cm.
a- Calculate the max mass that can be measured with this spring. Please explain your answer.
b- We suspend a weight of 120 N. Is Hooke's law still applicable? Is the length produced by this weight valid? Explain why?

I see you posted this question multiple times and the mods have combined them.

The spring constant for a linear spring is calulated from:


k = \frac {F}{\Delta x}


Here you know that the force of 20N extends the spring by 2 cm. so you can determine the value of k. Then given that the max allowable \Delta x is 10 cm, use F=k \Delta x to find the max force that doesn't exceed the 10cm allowable extension.

As for the second part of the question, what do you tghink will happen if the max allowable linear extension is exceeded?

Thank you for your helping me answer the third problem. Can you please help me answer the first two problems?

A spring is shortened to 2 cm by a force of 0.6N
a- Calculate the spring constant of the spring.

Use the same process as I outlined earlier.


b- What length variation produces a piece of iron of 200 \(cm^3\) volume?
c- What is in this case its final length (p iron = 7800 kg/ \(m^3\)?

Sorry, but this makes no sense. Something has been lost in translation.

A spring has a length of 40 cm at rest, we hang to its free extremity an empty bottle, its length becomes 42 cm, we fill the bottle by 1.5 liter, the final length of the spring is 50 cm.
What is the weight of the empty bottle? (p pure alcohol = 800 kg/ m^3. Please explain your solution in details.

The change in length from 42 to 50 cm is due to the weight of 1.5L of alcohol. You need to determine the weight of the alcohol and then calculate the spring constant. Then you can determine the weight of the empty bottle from the fact that the spring stretched 2 cm due to the bottle alone.

This is the original questions for Problem 1 (b, c)
A spring is shortened to 2 cm by a force of 0.6N
b- What length variation produces a piece of iron of 200 cm^3 volume?
c- What is in this case its final length (p iron = 7800 kg/m^3)?
I am not sure what is (p iron = 7800 kg/m^3) ans what we need it for!

I think I see - you need to determine the weight of a piece of iron that is 200 cm^3 in size, given density of iron (using greek symbo rho: \rho for density) of 7800 Kg/m^3, then determine how much the spring extends based on the spring value you calcuated in part 'a.'. In other words parts 'b' and 'c' is one question, not two.