Elastic Potential Energy and the Spring Constant

[Soaring Crane]

1) A 60 kg person drops from rest a distance of 1.20 m to a platform of negligible mass supported by a stiff spring. The platform drops 6 cm before the person comes to rest. What is the spring constant of spring?

a.4.12E5-------b. 2.56E5---------c. 3.92E5--------d. 5.45E4----e. 8.83E4


I don’t know if I did this setup correctly, but I got c. as my answer. Please tell me if my equations are incorrect or if I neglected another energy quantity.

U2 = U1

Mgh2 + (k*x_2^2)/2 = mgh1 + (k*x_1^2)/2
(k*x_2^2)/2 = mgh1

k = (2mgh)/(x^2) = (2*9.80 m/s^2*1.20 m)/(0.06 m)^2 = 392000 N/m

2) A spring-loaded dart gun is used to shoot a dart straight up into the air, and the dart reaches a maximum height of 24 m. The same dart is shot up a second time from the same gun, but this time the spring is compressed only half as far (compared to first hot). How far up does dart go this time (neglect friction and assume spring obeys Hooke’s law)?

a.48 m---------b. 12 m---------c. 6 m------------d. 3m

I followed the same format as in #1, so, if that is wrong, this would be, too.

(k*x_2^2)/2 = mgh

at 24 m, h = (k*x^2)/(2mg) = 24 m

when x = x/2, h = (k*x^2)/(4*2 mg) = (1/4)* (k*x^2)/(2mg) = (1/4)(24 m) = 6 m ?



Thanks.

 

[radou]

Looks good to me, I hope I'm not missing something.

 

[kyle8921]

Your equations and setup are correct for both problems. For the first problem, the answer is indeed c. 3.92E5, and for the second problem, the answer is c. 6 m. Great job using the equations for elastic potential energy and Hooke's law to solve these problems! It's important to note that in these types of problems, we are assuming that all the energy is conserved and there is no loss due to friction or other factors. Keep up the good work!