Work of friction on a moving box

[igotserv3d]

A box of mass is sliding along a horizontal surface.

The box leaves position with speed . The box is slowed by a constant frictional force until it comes to rest at position .

Find , the magnitude of the average frictional force that acts on the box. (Since you don't know the coefficient of friction, don't include it in your answer.)

F_friction =(1/(2*x_1))*(m*v_0^2) <- we found this part


Part B
After the box comes to rest at position x1, a person starts pushing the box, giving it a speed .

When the box reaches position x2 (where x2>x1 ), how much work W_p has the person done on the box?

Assume that the box reaches x2 after the person has accelerated it from rest to speed x1.
Express the work in terms of M, V_0, V_1, x_1, and x_2

Assume that the box reaches v_2 after the person has accelerated it from rest to speed v_1.

W_p (work of the person) = ?



we are having a problem including friction in the equation since we can't use mu.

We figure that...
work of the person = delta_k + (work of friction)

does anybody know how to find the work of friction without mu? or by using its change in kinetic energy.

 

[marlon]

Use $$W = E_k^{END}-E_k^{BEGINNING} = d*F$$

D is the distance traveled in the direction of the friction force F. You don't need to know mu in order to find F. Just apply the above theorem

marlon

 

[thepatientmental]

To find the work of friction without using the coefficient of friction, we can use the formula W_friction = F_friction * d, where F_friction is the average frictional force and d is the distance traveled. We already know the distance traveled (x1) and the mass of the box (m), so we just need to find the average frictional force.

From the given information, we know that the box starts with an initial speed v0 and comes to rest at position x1. This means that the change in kinetic energy (delta_k) is equal to the initial kinetic energy (1/2 * m * v0^2). So, we can rewrite the formula for the work of the person as:

W_p = (1/2 * m * v0^2) + W_friction

Now, to find the work of friction, we can substitute the formula for W_friction into the equation:

W_p = (1/2 * m * v0^2) + (F_friction * x1)

We already know the initial speed (v0) and the mass of the box (m), so we just need to find the average frictional force (F_friction). To do this, we can use the formula we found in Part A:

F_friction = (1/(2*x1)) * (m * v0^2)

Substituting this into the equation for W_p, we get:

W_p = (1/2 * m * v0^2) + ((1/(2*x1)) * (m * v0^2) * x1)

Simplifying, we get:

W_p = (1/2 * m * v0^2) + (1/2 * m * v0^2)

Finally, we can rewrite this in terms of the given variables:

W_p = (1/2 * M * V_0^2) + (1/2 * M * V_0^2)

Therefore, the work of the person is equal to the initial kinetic energy (1/2 * M * V_0^2) plus the work of friction (1/2 * M * V_0^2). This result makes sense, as the person's work is equal to the change in kinetic energy of the box (from rest to speed x1) plus the work done by friction to slow down the box.