Work of friction on a moving box

  • Thread starter igotserv3d
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  • #1
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.
 

Answers and Replies

  • #2
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Use [tex]W = E_k^{END}-E_k^{BEGINNING} = d*F[/tex]

D is the distance travelled 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
 

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