A 250g wood block firmly attached to a horizontal spring slides along a table with a coefficient of friction of 0.40. A force of 30.N compresses the spring 20cm. If the spring is released from this position what will be the block's speed when it passes equilibrium position?

Answer 1

k = 30 N/.2 m = 150 N/m

PE of spring = (1/2) k x^2 = 75(.04)=3 Joules

normal force = m g
friction force = .4 m g
work done against friction = .2 *.4 * mg
= .08*.25*9.81 = .196 Joules

so Ke left = 3 - .196 = 2.8 Joules
(1/2) m v^2 = 2.8
calculate v