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
[tex]-(\frac{1}{2} kd^{2} )-(fk*m*g*d^{2} )=\frac{1}{2} m(v^{2} -v_{i} ^{2} )[/tex]
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:
[tex]-(\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[/tex]
