Varying coefficient of friction

[eeriana]

Homework Statement

A box is sliding with a speed of 4.50 m/s on a horizontal surface when, at point P, it encounters a rough section. On the rough section, the coefficient of friction is not constant, but starts at .100 at P and increases linearly with a distance past P, reaching a value of .600 at 12.5m past point P. a) Use the work energy theorem to find how far this
box slides before stopping.

Homework Equations

Work = .5mV1-.5mV2
K= .5mv^2

The Attempt at a Solution

I am not really sure where to begin, if someone could steer me in the right direction, I would be grateful. I know that the friction area will cause a deceleration until the box stops. I know that the net force of Fx will be the F force in the positive x direction minus the Friction force in the negative x direction. I feel like I need the mass of the box and since it is not given, it must not be necessary and that means I am missing something.
I believe the s vector will be 12.5 since that is the distance of the friction area.

Thanks..

Amy

 

[Astronuc]

Since the friction force is related to the coefficient of friction $\mu$, one can use the fact that the kinetic energy is dissipated by the friction over the distance from P to the point that it stops, where the kinetic energy is of course zero. The friction force is simply $\mu$(x)*mg.

So the initial KE = 1/2 mv² = energy dissipated (or work done by friction). Note the mg is constant, but the friction coefficient is a function of x or the distance traveled.

 

[fajoler]

Alright so I did relate that Work is equal to the change in kinetic energy which is also equal to the magnitude of Force times the magnitude of the change in distance.By doing so, I get that the initial Kinetic Energy equals mg(0.04x + 0.1)xThe equation for u(x) = 0.04x + 0.1. The distance traveled during the process is x.When I solve for x, I get the answer 3.984 m, but this seems to be the wrong answer. Does anyone see where I went wrong?

 

[fajoler]

bump*

 

[rwisz]

,

I would approach this problem by first defining the variables and equations that are relevant to the situation. In this case, the key variables are the coefficient of friction (μ), the initial speed (v1), the final speed (v2), and the distance (s) traveled by the box. The relevant equations are the work-energy theorem, which relates the work done on an object to its change in kinetic energy, and the equation for kinetic energy, which relates an object's mass, velocity, and kinetic energy.

With these in mind, we can write the work-energy theorem as W = ΔK = 0.5mv2^2 - 0.5mv1^2, where W is the work done by the friction force and ΔK is the change in kinetic energy of the box. We can also write the equation for kinetic energy as K = 0.5mv^2, where K is the kinetic energy, m is the mass of the box, and v is its speed.

Next, we can use the definition of work as the product of force and distance, W = Fd, to express the work done by the friction force as W = μmgd, where μ is the coefficient of friction, m is the mass of the box, g is the acceleration due to gravity, and d is the distance traveled by the box. We can also use the definition of average velocity, v = Δx/Δt, to express the change in kinetic energy as ΔK = 0.5mΔv^2 = 0.5m(v2^2 - v1^2).

Now, we can substitute these expressions into the work-energy theorem to get μmgd = 0.5m(v2^2 - v1^2). We can then rearrange this equation to solve for the distance traveled by the box, d = 0.5(v2^2 - v1^2)/(μg). Plugging in the given values of v1 = 4.50 m/s, v2 = 0 m/s (since the box stops at the end of the rough section), and μ = 0.100, we get d = 0.5(0 - 4.50^2)/(0.100*9.8) = 9.18 m.

Therefore, the box will slide a distance of 9.18 m on the rough