How Far Will a Box Slide Before Stopping If Pushed with a Force of 21.9N?

[Alem2000]

a stockroom worker pushes a box with mass 11.2kg on a horizontal surface with a constant speed of 3.50m\s^2 The coeficient of friction is (for surface of the ground and box) .20 A...what horizontal force must the worker apply to maintian the motion.. i got $$F_x=21.9N$$ which is correct and then it asks me "it the force calculated in part a) is removed, how far does the box slide before coming to rest?" Where do i start..all I have is velocity?

 

[TenaliRaman]

Once the applied force is removed,
what is the net force?
isn't net force = ma?
does that mean we have deceleration?
oh do we know initial velocity?
and what abt final velocity?
can i get the time?
hmm but s = ut + 1/2at^2 ?
so?
ah yes!

-- AI

 

[wais]

To calculate the distance the box will slide before coming to rest, we can use the equation for constant speed motion: d = vt, where d is the distance, v is the velocity, and t is the time.

In this case, we know the velocity (3.50m/s) and we can calculate the time using the formula for acceleration: a = F/m, where a is the acceleration, F is the force, and m is the mass.

Since we know the acceleration (0m/s^2, since the box is moving at a constant speed), the mass (11.2kg), and the force (21.9N, as calculated in part a), we can rearrange the formula to solve for time: t = m/a = 11.2kg / (21.9N / 11.2kg) = 11.2kg / 1.95m/s^2 = 5.74s.

Now, we can plug this time into the distance formula to find the distance the box will slide before coming to rest: d = vt = (3.50m/s) * (5.74s) = 20.09m.

Therefore, the box will slide 20.09 meters before coming to rest if the force calculated in part a) is removed.