Question

A box of mass m is sliding along a horizontal surface.Part A) The box leavesposition x = 0 with speed v0. The boxis slowed by a constant frictional force until it comes to rest at position x = x1.Find the magnitude of the average frictional force thatacts on the box. (Since you don't know the coefficient of friction,don't include it in your answer.) Express thefrictional force in terms of m, v0, x1.

Answer 1

Answer:

$f=\dfrac{mv_0^2}{2x_1}$

Explanation:

Given:

Initial position of the box is, $x=0$

Final position of the box is, $x=x_1$

Initial speed of the box is $v_0$

Final speed of the box is 0 as it comes to rest.

Now, displacement of the box is given as the difference of final and initial positions. So,

Displacement = Final position - Initial position

$\Delta x=x_1-0=x_1$

Now, from the definition of work-energy theorem, we know that, the work done by the net force is equal to the change in the kinetic energy of the body. So, as per this theorem;

Work done by friction = Change in kinetic energy of the box.

Work done by a body is equal to the product of force and displacement caused along the line of application of force. If the force and displacement are in same direction, work is positive and if both force and displacement are opposite to each other, then work is negative.

Here, friction acts in the direction opposite to displacement. So, work done by friction is negative.

Work done by friction =Frictional force × Displacement

$W=-fx_1$

Now, change in kinetic energy is given as:

$\Delta KE=Final\ KE-Initial\ KE\\\\\Delta KE=\frac{1}{2}m(0)^2-\frac{1}{2}mv_0^2\\\\\Delta KE = -\frac{1}{2}mv_0^2$

Therefore, $W=\Delta KE$

$-fx_1=-\frac{1}{2}mv_0^2\\\\f=\frac{mv_0^2}{2x_1}$

Therefore, frictional force in terms of 'm', 'v₀' and 'x₁' is given as:

$f=\dfrac{mv_0^2}{2x_1}$