a box slides

[logistic_guy]

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 $\displaystyle 0.25$ and the push imparts an initial speed of $\displaystyle 2.5 \ \text{m/s}$?

 

[khansaheb]

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 $\displaystyle 0.25$ and the push imparts an initial speed of $\displaystyle 2.5 \ \text{m/s}$?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem

 

[logistic_guy]

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 $\displaystyle 0.25$ and the push imparts an initial speed of $\displaystyle 2.5 \ \text{m/s}$?

If you push a box with a force $\displaystyle F$ and there is friction on the floor, the sum of the forces are:

$\displaystyle F - F_R = ma$

If you remove your hand from the box, your force $\displaystyle F$ is no longer acting on the box, so the forces become:

$\displaystyle -F_R = ma$

Or

$\displaystyle -\mu mg = ma$

Or

$\displaystyle -\mu g = a \ \ \ \ \textcolor{red}{(1)}$

where $\displaystyle \mu$ is the coefficient of friction.

If we assume that the acceleration $\displaystyle a$ is constant, we can write it in terms of the distance $\displaystyle x$ from this equation:

$\displaystyle v^2 = v^2_0 + 2ax$

We know that the box will stop at some moment so the final velocity $\displaystyle v = 0$.

$\displaystyle 0 = v^2_0 + 2ax$

This gives:

$\displaystyle a = -\frac{v^2_0}{2x}$

Plug this in equation $\displaystyle \textcolor{red}{(1)}$

$\displaystyle -\mu g = \left(-\frac{v^2_0}{2x}\right)$

Or

$\displaystyle \mu g = \left(\frac{v^2_0}{2x}\right)$

Plug in numbers.

$\displaystyle 0.25(9.8) = \left(\frac{2.5^2}{2x}\right)$

This gives:

$\displaystyle x = \textcolor{blue}{1.28 \ \text{m}}$