Can Friction Actually Speed Up an Object?

[jchojnac]

Homework Statement

It is sometimes claimed that friction forces always slow an object down, but this is not true. If you place a box of mass M on a moving horizontal conveyor belt, the friction force of the belt acting on the bottom of the box speeds up the box. At first there is some slipping, until the speed of the box catches up to the speed v of the belt. The coefficient of friction between box and belt is . Do not worry about italics. For example, if a variable g is used in the question, just type g and for use mu.

(a) What is the distance d (relative to the floor) that the box moves before reaching the final speed v? (Use energy arguments to find this answer.)

(b) How much time does it take for the box to reach its final speed?

Homework Equations

W=-Fd/2
K=1/2mv^2
f=mu*N

The Attempt at a Solution

Can't figure it out.

 

[tiny-tim]

Welcome to PF!

Hi jchojnac! Welcome to PF!

(have a mu: µ )

If you place a box of mass M on a moving horizontal conveyor belt, the friction force of the belt acting on the bottom of the box speeds up the box. At first there is some slipping, until the speed of the box catches up to the speed v of the belt. The coefficient of friction between box and belt is … mu.

(a) What is the distance d (relative to the floor) that the box moves before reaching the final speed v? (Use energy arguments to find this answer.)

Hint: for (a), use the work-energy theorem:

work done by friction = energy gained

 

[nlightner]

I would like to clarify that the statement "friction forces always slow an object down" is not entirely accurate. While friction does often act to slow down an object, it is not always the case. In certain situations, like the one described in the problem, friction can actually speed up an object.

To solve for the distance d, we can use the principle of conservation of energy. Initially, the box has no kinetic energy and all the work done on it by the friction force is converted into potential energy. As the box speeds up, the potential energy is converted into kinetic energy. Therefore, we can equate the work done by the friction force with the change in kinetic energy:

W = Kf - Ki

Where W is the work done by the friction force, Kf is the final kinetic energy, and Ki is the initial kinetic energy (which is zero).

We can also express work done by the friction force as the product of the friction force and the distance d:

W = f*d

Substituting this into our equation, we get:

f*d = Kf - Ki

Since f = mu*N, we can rewrite this as:

mu*N*d = 1/2*M*v^2

We know that N = Mg, so we can further simplify to:

mu*M*g*d = 1/2*M*v^2

Canceling out the mass and rearranging, we get:

d = v^2 / (2*mu*g)

This is the distance that the box moves before reaching the final speed v.

To solve for the time it takes for the box to reach its final speed, we can use the equation for average velocity:

v = d/t

Substituting in our previously calculated value for d, we get:

v = (v^2 / (2*mu*g))/t

Solving for t, we get:

t = 2*mu*g/v

This is the time it takes for the box to reach its final speed v.