Please solve this for me: The figure shows a rock with the mass of 8.00kg that l
ID: 2175563 • Letter: P
Question
Please solve this for me:
The figure shows a rock with the mass of 8.00kg that lays still on the top of the spring.
The rock is NOT attached to the surface it lays on.
In this position the spring compressed 10.0cm comparatively its equilibrium length.
a) What is the size of the springconstant k?
The rock is pressed downward 30.0cm and is "released".
b) What is the potential energy in the compressed spring immediately before the system is released?
c) What is the change in the potential energy assosiated with gravity, when the rock moves FROM the release point TO the point it momentarily comes to rest in maximum height above the ground?
d) What is the actual value for maximal height measured from the release point ?
Please help me!!!
Explanation / Answer
a) mg = kx
=> k = mg/x = 9.81*8/.1= 784.8 N/m
b) Potential energy in the spring = 1/2kx2 = 1/2 * 784.8 * 0.4*0.4 = 62.784 J
c)
Energy of the system is conserved
62.784 + mg*0.4 = mgh where h = height from the uncompressed position of spring ( 40 cm above release point)
=>Change in energy = 62.784 J
mg(h-0.4) = 62.784
=> h-0.4 = 0.8
=> h =120 cm
Hence from the release point the stone goes 160 cm up..
Hope this will do. :)