What is the velocity of the toy head as it moves in the vertical direction?
If the spring of a jack in the box is compressed a distance of .08m from its relaxed length and then released, what is the speed of the toy head when the spring returns to its natural length? Assume the mass of the toy head is .05kg, the spring constant is 80M/m, and the toy head moves only in the vertical direction. Also disregard the mass of the spring.
(Hint: remember that there are two forms of potential energy in this problem)
Solve for m/s.
Please show work.
Best answer:
Answer by civil_av8r
Consevation of energy
Initial Energy = Final Energy
Initial Spring Energy + Initial PE + Initial KE = Final Spring Energy + Final PE + Final KE
In this case
vi = 0
hi = -.08
xi = -.08
vf =?
hf = 0
xf = 0
1/2*k*xi^2 + m*g*hi = 1/2*m*vf^2
Solve for vf
vf = sqrt((k*xi^2 + 2*m*g*h)/m) = 2.94 m/s
So, the -.08 term will become positive when squared, and h will still be negative, which makes since, because the spring has to push the jack against gravity.
What do you think? Answer below!
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Hai Jon, what you have said is very true. Initially the energy is of elastic energy + 0 gravitational PE+0 KE
ie 1/2 k x^2 ie 1/2*80*(0.08)^2 = 0.256 J
When the toy head is at the normal length position, it has KE and PE and 0 elastic energy.
Let V be the velocity. So KE is 1/2 * 0.05* V^2 and PE is mgh = 0.05*9.8*0.08 = 0.0392 J
Hence the total energy = 0.025V^2 + 0.0392 J
By conservation of energy, 0.025V^2 + 0.0392 = 0.256
Simplifying we get V^2 = 8.672
V = 2.945 m/s