Question

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.

Answer 1

Answer:

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

$-(\frac{1}{2} kd^{2} )-(fk*m*g*d^{2} )=\frac{1}{2} m(v^{2} -v_{i} ^{2} )$

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:

$-(\frac{1}{2} *200*0.1^{2} )-(0.15*2*9.8*0.1^{2} )=\frac{1}{2} *2*(v^{2} -4^{2} )\\-1-0.0294=v^{2} -16\\v=\sqrt{-1-0.0294+16} =3.87m/s$