A 2.0-kg block sliding on a rough surface is attached to one end of a horizontal spring (k = 200 N/m) which has its other end fixed. If the block has a speed of 4.0 m/s as it passes through the equilibrium position, what is its speed when it is 10 cm from the equilibrium position? The coefficient of friction between the block and surface is 0.15.

The speed is 3.87 m/s

Explanation:

The total work is equal to the change to kinetic energy:

Wtotal = ΔEk

Wspring + Wfriction = Ekf - Eki

Eki = 0

Where

k = 200 N/m

d = 10 cm = 0.1 m

fk = 0.15

m = 2 kg

g = 9.8 m/s²

vi = 4 m/s

Replacing and clearing v:

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explanation:

trig 15.3 cos