Question

An elastic spring constant of 150 N / m was compressed by a block 15 cm in 5.0 kg cm. Determine the speed of the block when leaving the spring, knowing he is in a horizontal plane.

Answer 1

Spring
F = -kx
PE = (1/2) k x^2 = (1/2)(150 N/m)(.15m)^2 = 1.6875 J

PE becomes KE of block = (1/2) m v^2.
Knowing PE = KE and m, find v.

Answer 2

I marked the letter

a)

Correct?

Answer 3

@douglaswinslowcooper

Answer 4

1.68 J = (1/2)(5 kg)(v^2)

v^2 = 0.672
v = 0.82 m/s correct. Well done.

Answer 5

thank you!!!