Finding the Maximum Compression Distance of a Spring: A Physics Problem

[roam]

Homework Statement

A 2kg block is dropped from a height of 40 cm onto a spring of force constant k=1960 N/m. Find the maximum distance the spring is compressed.

(The answer must be 10 cm)

The Attempt at a Solution

Well, I know that if a spring is stretched "y", it will be compressed "y". The problem is that I can't find out how long will the spring stretch once it is dropped from that height. Therefore I tried to use the formula

$$F = ky$$

Where $$F= mg = 2 \times 9.81$$ and k =1960 N/m which is 196000 N/cm

$$2 \times 9.81 = (196000) k$$

$$y= \frac{2 \times 9.81}{196000}$$

But this gives me the wrong answer. Why is that??

By the way, I know that the tension is $$T= 2 \pi \sqrt{\frac{m}{k}} = 2 \pi \sqrt{\frac{2}{196000}}= 0.02 N$$.

 

[rock.freak667]

It is best to consider energy for this problem, graviational pe = potential pe of springAlso that T is period time and not the tension.

 

[roam]

It is best to consider energy for this problem, graviational pe = potential pe of spring

I tried that, it doesn't seem to work:

$$mgh= \frac{1}{2} k x^2$$

2 (9.81) 40 = 1/2 (196000) x²

x=0.089 cm

Because the correct answer should be 10 cm! Is there anything wrong with my calculations?

Also that T is period time and not the tension.

Yep, my mistake.

 

[rock.freak667]

Not sure, the only way to get 10cm exact is if the mass was dropped from 50cm.

Normally, the energy method should work.

 

[rdl]

I would suggest approaching this problem using the conservation of energy principle. When the block is dropped from a height of 40 cm, it will have a potential energy of mgh = 2 x 9.81 x 0.4 = 7.84 J. This energy will be transferred to the spring when it compresses, causing it to have a potential energy of 1/2 kx^2, where x is the maximum distance the spring is compressed. Setting these two energies equal to each other, we get 7.84 = 1/2 x 1960 x x^2, which simplifies to x = 0.1 m or 10 cm. This approach takes into account the initial potential energy of the block and the potential energy stored in the compressed spring, giving us a more accurate result.