Frictional forces of pushing a box

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Homework Help Overview

The discussion revolves around a physics problem involving frictional forces as a worker pushes a box on a horizontal surface. The box has a specified mass and is moving at a constant speed, with the problem requiring the calculation of forces involved in maintaining motion and subsequent motion after the applied force is removed.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between applied force and friction when the box is in motion. There is an exploration of how to determine the box's acceleration and distance traveled after the applied force is removed. Some participants question the calculations related to acceleration and the application of kinematic equations.

Discussion Status

Several participants have provided insights into the problem, particularly regarding the transition from constant motion to deceleration due to friction. There is an ongoing exploration of kinematic methods to find the distance traveled after the force is removed, with some participants sharing their attempts and results.

Contextual Notes

Participants are navigating the implications of removing the applied force and the resulting dynamics of the box, including the effects of friction and the need for accurate sign conventions in their calculations.

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Frictional forces [SOLVED]

Homework Statement


A stockroom worker pushes a box with mass 11,2 kilograms on a horizontal surface with a constant speed of 3.5 m/s. The coeffisient of kinetic fricion between the box and the surface is 0.2.

In a) i had to find the force that the worker must apply to maintain the motion.

b) If the force calculated in part a) is removed, how far does the box slide before coming to rest.


Homework Equations





The Attempt at a Solution


I am pretty sure that I solved task a). I found the frictionforce to be 22N, and since the box is having a constant speed, the sum of forces in x-direction have to be zero.
Force = Frictionforce in this case.

But i really need some tips on the B part. Anyone?

Thanks!
 
Last edited:
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Once you remove the applied force of the push, the net force is no longer zero: You have the unbalanced friction force acting to slow the box down. Find the acceleration and use kinematics to find the distance. Or use energy methods.
 
Well the force that was being applied only acted to keep the box moving at 3.5m/s so when it is removed the box will start deccelerating from this initial speed.

You need to find the acceleration of the box due to the frictional force, which is dependent on the boxes mass. With this you can determine the distance the box travels.
 
Thanks for fast replies:) Ok, I tried something like that, but I didn't come out with an correct answer. When the force is removed I have:

Force-Friction = ma

0 - Friction = ma

a = Friction / m = -1.96m/s^2

But how do I find the distance now?
 
It's a kinematics problem now. Accelerated motion.
 
Thanks

Doc Al said:
It's a kinematics problem now. Accelerated motion.

I did of course make a mess of things with the sign. Got the distance to be 3,1 meters after finding the time 1.78s.
 
Sounds good.
 

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