How Far Does a Box Slide Before Stopping If Pushing Force Is Removed?

[Chandasouk]

Homework Statement

A stockroom worker pushes a box with mass 11.1 kg on a horizontal surface with a constant speed of 3.80 m/s. The coefficient of kinetic friction between the box and the surface is 0.170.

Homework Equations

F=ma
w=mg


What horizontal force must be applied by the worker to maintain the motion?

I found that the Fapplied is 18.5N


If the force calculated in part A is removed, how far does the box slide before coming to rest?

This question I'm having problems on. Here is what I did

18.5N = 11.1kga

a=1.66666667 m/s^2

Then I used the equation of

Vfinal^2=Vinitial^2+2a$$\Delta$$x

$$\Delta$$x = 0 - 14.44m/s
--------------
2(1.66666667)


$$\Delta$$x=4.3m ?

 

[rock.freak667]

Yes that is correct.

Also you could have used conservation of energy.

Note that your a=1.66666667 m/s² should really be a=-1.66666667 m/s²

But I believe you put it as negative while working it out, so just remember to write the - sign else your teacher might give you it wrong.

 

[tomcue]

I would like to first clarify that the calculations and equations used in this response are correct. The horizontal force required to maintain the motion of the box is indeed 18.5N, and the distance the box will slide before coming to rest is approximately 4.3 meters.

To further explain, the force of friction acting on the box is equal to the coefficient of kinetic friction (μ) multiplied by the normal force (mg), where m is the mass of the box and g is the acceleration due to gravity. In this case, the normal force is equal to the weight of the box, which is mg. So, the force of friction can be calculated as μmg.

In order to maintain a constant speed, the applied force must be equal and opposite to the force of friction. So, Fapplied = μmg. Plugging in the given values, we get Fapplied = 0.170 * 11.1 * 9.8 = 18.5N.

Now, if we remove this force, the only force acting on the box is the force of friction, which will cause the box to decelerate. Using the equation ∆x = (v^2 - u^2)/2a, where v is the final velocity (0 m/s), u is the initial velocity (3.80 m/s), and a is the deceleration (1.66666667 m/s^2), we can calculate the distance the box will travel before coming to rest. Plugging in the values, we get ∆x = (0^2 - 3.80^2)/2(-1.66666667) = 4.3 meters.

So, to answer the question, if the applied force is removed, the box will slide approximately 4.3 meters before coming to rest. This is due to the deceleration caused by the force of friction, which is acting in the opposite direction of the motion.