A 2-kg block falls from rest a vertical distance of 0.75 meters above the top of a spring. The block lands on top of the spring compressing it 0.35 meters from its equlibrium position.
a) What is the gravitational potential energy of the 2-kg mass just before it is dropped above the spring?
b) Determine the speed of the mass at the instant the mass makes contact with the uncompressed spring.
c) Using the conservation of energy, calculate the spring's spring constant.
d) If you could place a 1-kg mass on the same spring and could push down compressing the spring 0.50 meters from its equlibrium position, how high would a 1-kg mass be fired vertically if you released the compressed spring?
PE (Gravitational) = m*g*h
PE (Spring) = (1/2)*(spring constant)*(distance compressed)
KE = (1/2)(m)(v^2)
The Attempt at a Solution
I first tried to start part a by using my first equation.
PE = (2 kg)(9.81 m/s^2)(.75 m)
PE = 14.715 J
Now my instructor told me that the answer to this part was 21.6 J so I already got something wrong here.
Anyway, for part b:
I figured that the gravitational potential energy of the block before it was dropped (the answer to part a) was equal to the kinetic energy at the instant the mass makes contact with the uncompressed spring. Thus, PE (g) = KE.
14.715 J = (1/2)(2 kg)v^2
My instructor said that the answer was 4.65 m/s. If I had used 21.6 J as the total KE, then I got that correct answer. So, therefore, my mistake must be somewhere in part a.
(I got parts c and d both correct but only when I utilized the 21.6 J that I was supposed to get in part a.)
Does anyone know how to get the 21.6 J for part a? Thanks. That's really the only issue in this problem.
Also, my instructor said that he did these answers by himself rather quickly, so there's a minor chance of a careless error he made. But I don't think that's the case.