Finding the Final Velocity of a Box Sliding Across a Rough Surface

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Homework Help Overview

The discussion revolves around determining the final velocity of a box sliding across a rough surface after experiencing friction. The initial speed of the box is given as 3.8 m/s, and it encounters a rough region with a coefficient of friction of 0.7 over a distance of 0.52 m. Participants are exploring the application of the work-energy theorem in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to apply the work-energy theorem to relate the forces acting on the box to its change in kinetic energy. There is discussion about the sign of the work done by friction and how it affects the final velocity calculation. Some participants question the assumptions made regarding the direction of forces and the resulting equations.

Discussion Status

Multiple interpretations of the work-energy theorem are being explored, particularly regarding the sign of the work done by friction. Some participants suggest that the final velocity should logically be less than the initial velocity, prompting a reevaluation of the equations used. Guidance has been offered regarding the negative work done by friction, but no consensus has been reached on the correct approach.

Contextual Notes

Participants are grappling with the implications of the frictional force acting against the motion of the box, and there is an acknowledgment that the calculations must reflect this negative work. The discussion is ongoing, with no definitive resolution yet.

taveuni
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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?
 
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taveuni said:
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

Because the force of friction is doing work against the motion, it should actually be:

-F*d= .5mvf^2 - .5*m*vi^2

When calculating work, it's force*distance*cos(theta), where theta is the angle between the force and the movement. In this case, theta is 180 degrees, so cos(theta) = -1
 
Thank you so much.
 
taveuni said:
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)

Ridiculous. What am I doing wrong?

You must realize that, by definition, F.d is negative. The force is not in the direction of motion.

EDIT: Guess I was a bit late. ;)
 
I'm going to venture a guess and say that since you know that logically the object will slow down, vf will be less than vi, so in the work-energy equation you have, the right hand side would have to be negative to satisfy this condition. Hence the left hand side would have to be negative to make satisfy the equality. Also, since work = F*d*cos(theta) and theta here is 180 degrees since it is acting in the direction opposite of the motion. I'm not too sure on any of this, but when i put in a negative 3.43 i get a final velocity around 2.something m/s, which makes more sense

EDIT: Guess I was even later...
 

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