A vertical spring (ignore its mass)?

Question

A vertical spring (ignore its mass), whose spring stiffness constant is 950 N/m, is attached to a table and is compressed down 0.150m. (a) an upward speed can it give to a 0.30 kg ball when released? (b) How high above its original position (spring compressed) will the ball fly?

Answer 1

This is a conservation of energy problem.

TE(C) = PE(C) = k delX; where k = 950 kg-m/sec^2//m and delX = .150 m.

TE(R) = KE(R) = 1/2 mV^2; where m = .3 kg and V = the initial upward velocity upon release.

The total energies are equal TE(C) = k delX = 1/2 mV^2 = KE(R); so that V = sqrt(2k delX/m) and you can do the math.

For the second half (b), set TE(R) = TE(h) = mgh = PE(h); where h is the height above the uncompressed spring (i.e., at point of release or launch). To get the height H above the compressed position we have H = h + delX. Thus we have TE(R) = 1/2 mV^2 = mgh = PE(h); so that h = V^2/2g and you found V in (a). Therefore H = V^2/2g + delX and you can do the math.

The physics is this...due to conservation of energy, the total energies TE are constant and equal at the three points in the trajectory of the ball: resting on the compressed spring where all the energy is the potential energy of the spring, right at release point where all the energy is the kinetic energy of the ball, and at the maximum height where all the energy is the location potential energy at height h.

Answer 2

Given:
K=950N/M
x=0.150m

Part A.
Ue= 1/2mv^2
Ue= (950N/m)(0.150m)^2
Ue=10.6875

Next plug it into the equation:
KE=1/2mv^2 the formula re-arranged gives us v=squareroot 2KE/m
10.688=1/2(0.30kg)v^2
v=sqrt of 2(10.688) / 0.30
v=8.4m/s

Answer 3

Use the formulation of ability capability saved interior the spring = kinetic capability of the ball a million/2 ok x^2 = a million/2 m v^2 ok x^2 = m v^2 v =x* sqrt(ok/m) = 0.one hundred ten * sqrt (820/0.500) = 4.455 m/s