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I think I solved the following problem but the answer (given by my professor) does not match with mine. Therefore, I would be happy if someone could confirm me that he is wrong ;)

There is a spring attached to a box of 200 gr. mass. The box is lifted up so that the lower end of the spring is 30 cm above the ground. Suddenly, the box is falling down to the ground. What is the spring's compression through the impact (s) ?

The spring's constant (k) is 156,96 N/m.

I went this way:

Duration of the fall:

h= 0.3 m

[tex] t= \sprt{\frac{2*h}{g}} = 0.247 s [/tex]

with the duration I get the final speed of the box:

[tex] v= gt = 2,43 \frac{m}{s} [/tex]

My thougt it that the potential energy of the spring has to be the amount of the kinetec energy and the potential energy before the impact:

[tex] \frac{1}{2}mV^2+mgh = \frac{1}{2}ks^s [/tex]

this equation gives me the compression (s) :

[tex] \sqrt{\frac{mV^2+2mgh}{k}}= 0.1225 m [/tex]

The answer given by my professor is 0,1 m.

Do you think he made a mistake as well ?

Thanks ! fara

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# Homework Help: I would like to get a confirmation about this one

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