Using Newton's Laws: Friction, Circular Motion, Drag Forces

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Sunwoo Bae
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Homework Statement
A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of kinetic friction is 0.15 and the push imparts an initial speed of 3.5m/s?
Relevant Equations
F-Ffr= m*a
v^2=(v0)^2+2*a(x-x0)
Here is my attempt at setting up the equation:
I set up the equation to find the acceleration of the box:
F-Ffr= m*a
after finding the acceleration, I can use the acceleration and plug it in the formula v^2=(v0)^2+2*a(x-x0), which will get me the value of (x-x0)The solution sheet says that F should be 0, and Ffr should be the only force acting on the box. However, I do not understand how F can be 0 when the box is given a push, and the push initiates the initial velocity of 3.5m/s. How can F be 0?
 
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Sunwoo Bae said:
I do not understand how F can be 0 when the box is given a push,
The motion occurred in two stages. First, the box was accelerated from rest by an unknown force and at an unknown acceleration. Second, the unknown force was terminated and the box allowed to decelerate under friction to a stop.
You are not asked anything about the first stage. You only have to consider the second stage, and in that stage F=0.