A 3.0 kg block is held against a spring with a force constant of 125 N/m. The spring is compressed by 12 cm. The ice is released across a horizontal plank with a coefficient of friction of 0.1
A) Calculate the velocity of the block just as it leaves the spring. Assume the friction between the plank and block is negligible until the moment the block leaves the spring.
B) Determine the distance the block travels after it leaves the spring.
Ek = 0.5mv^2
Ee = 0.5KX^2
W = Fdcostheta
The Attempt at a Solution
I did it just how I would do a problem except I threw in the work formula in there so:
Ek1 = Ee2 + (Fdcos180)
After cancelling the 0.5 in each equation: mv^2 = kx^2 + (fdcos180)
3v^2 = (125 x 0.12^2) - (0.1 x 3 x 9.81 x 0.12)
3v^2 = 1.45
V = 1.44 m/s
This is wrong because the answer key says the answer is 0.77 m/s