- #1
taveuni
- 16
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I am wondering if someone could guide me on this. I am getting a vf of 4.6m/s, which can't be right, considering vi is only 3.8m/s.
A box slides across a frictionless floor with an initial speed v = 3.8
m/s. It encounters a rough region where the coefficient of friction is
µk = 0.7. If the strip is 0.52 m long, with what speed does the box
leave the strip?
My thoughts:
F=µk*N=µk*g*m=6.867*m
vi=3.8 m/s
Using work-energy theorem:
F*d = .5*m*vf^2 - .5*m*vi^2
6.867*d*m = m*(.5*vf^2 - .5*vi^2)
m's cancel
6.867*.52 + .5*vi^2 =.5*vf^2
3.4335 + .5*(3.8)^2 =.5*vf^2
(3.4335 + 7.22)*2=vf^2
sqrt (21.307) = vf
4.62 = vf
Ridiculous. What am I doing wrong?
A box slides across a frictionless floor with an initial speed v = 3.8
m/s. It encounters a rough region where the coefficient of friction is
µk = 0.7. If the strip is 0.52 m long, with what speed does the box
leave the strip?
My thoughts:
F=µk*N=µk*g*m=6.867*m
vi=3.8 m/s
Using work-energy theorem:
F*d = .5*m*vf^2 - .5*m*vi^2
6.867*d*m = m*(.5*vf^2 - .5*vi^2)
m's cancel
6.867*.52 + .5*vi^2 =.5*vf^2
3.4335 + .5*(3.8)^2 =.5*vf^2
(3.4335 + 7.22)*2=vf^2
sqrt (21.307) = vf
4.62 = vf
Ridiculous. What am I doing wrong?