In a physics lab experiment, one end of a horizontal spring that obeys Hooke's law is attached to a wall. The spring is compressed 0.400 m, and a block with mass 0.300 kg is attached to it. The spring is then released, and the block moves along a horizontal surface. Electronic sensors measure the speed v of the block after it has traveled a distance d from its initial position against the compressed spring. The measured values are listed in (Figure 1) .
B)Use the workenergy theorem to derive an expression for v2 in terms of d.
Express your answer in terms of some or all of the variables k, m, d, μk and the acceleration due to gravity g.
1 Answer

The initial energy of the system is all in the spring:
Eo = Uo = ½kx² = ½k(0.400)²
for E and U in Joules and k in N/m.
Subsequently,
E = U + KE = Eo  work
½k(0.400  d)² + ½mv² = ½k(0.400)²  µmgd → multiply by 2
for d in meters, m in kg and v in m/s
k(0.160  0.800d + d²) + mv² = 0.160k  2µmgd
kd(d  0.800) + mv² = 2µmgd
mv² = kd(0.800  d)  2µmgd
v² = kd(0.800  d)/m  2µgd
so if by "v2" you meant "v²," that's your answer.
If you want the velocity, take the square root.
Hope this helps!
Source(s): plot: http://www.wolframalpha.com/input/?i=Plot{{0%2C+0}...