Chapter 4, Problem 40P

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 3.5 m/s?

To determine

The distance the box would travel before coming to rest when imparted with an initial speed by a force.

Answer to Problem 40P

Solution:

The box travels a distance of 4.2 mbefore coming to rest.

Explanation of Solution

When a box is given a push, such that it acquires an initial speed v , the box has kinetic energy. The box comes to rest due to the force of kinetic friction acting between the box and the floor, when it slides on the floor. The work done against the force of kinetic friction reduces the speed of the box to zero and it comes to rest.

Given:

Coefficient of kinetic friction between the block and the floor ${\mu }_k=0.15$

Initial speed of the box $v=3.5\text{ m/s}$

Formula used:

Kinetic Frictional force is obtained by:

  $F_{fr}={\mu }_k{F}_N$

The normal force would be equal to the weight.

  $\begin{array}{l}{F}_N=mg\\ \Rightarrow F_{fr}=\mu mg\end{array}$

Change in kinetic energy is obtained by:

  $K=\frac{1}{2}m{v}^2$

The work done would be equal to the change in kinetic energy.

  $\begin{array}{c}\frac{1}{2}m{v}^2=F_{fr}\cdot s\\ ={\mu }_{k}mg\cdot s\end{array}$

  $s=\frac{v^2}{2{\mu }_{k}g}$

Here, the coefficient of kinetic friction is ${\mu }_k$ , the force of kinetic friction is $F_{fr}$ , the normal force is $F_N$ , acceleration due to gravity is g , the mass of the box is m, its speed is v and the distance it travels before coming to rest is s .

Calculation:

Substitute the given values of v , ${\mu }_k$ and 9.8m/s² for g in the expression and solve.

  $\begin{array}{c}s=\frac{v^2}{2{\mu }_{k}g}\\ =\frac{{\left(3.5\text{ m/s}\right)}^2}{2\left(0.15\right)\left(9.8{\text{ m s}}^{-2}\right)}\\ \approx 4.2\text{m}\end{array}$

Conclusion:

When a box is pushed with an initial speed of 3.5 m/s on a floor with the coefficient of kinetic friction between the box and the floor being 0.15, it comes to rest after travelling a distance of 4.2 m.