Question

A block pushed along the floor with velocity $v0x$slides a distance d after the pushing force is removed.

a. If the mass of the block is doubled but its initial velocity is not changed, what distance does the block slide before stopping?

b. If the initial velocity is doubled to$2v0x$ but the mass is not changed, what distance does the block slide before stopping?

Short answer

(a) The mass of an object has no bearing on the object's distance travelled. As a result, even if the mass of the object is doubled, the distance travelled remains constant.

(b) If the original velocity is twice, the object's distance travelled is four times that of the starting value.

Step-by-step solution

Introduction

Kinetic friction is the friction that occurs when an object moves in the opposite direction of its motion. When a pushing force is removed from an object, the body comes to a complete stop due to kinetic friction.

Explanation (a)

Apply kinematic equations to an object's motion.

$v_{\mathrm{f}}^2-v_i^2=2as$

we see$v_{\mathrm{f}}$is velocity of the object, $v_i$is the starting velocity of the object, $a$is the acceleration of the object, and $s$is the distance travelled.

The object's frictional force acts in the opposite direction of its motion. The following is the kinetic frictional force acting on it:

$f_{\mathrm{k}}={\mu }_{\mathrm{k}}mg$

We see, ${\mu }_{\mathrm{k}}$is the coefficient of kinetic friction, $m$is the mass of the object, and $g$is the acceleration due to gravity.

The net force acting on it is calculated using Newton's second rule of motion:

$F_{\text{net}}=-f_{\mathrm{k}}$

$ma=-{\mu }_{\mathrm{k}}mg$

$a=-{\mu }_{\mathrm{k}}g$

Interchange and put 0 for $v_{\mathrm{f}},v_0$for $v_{\mathrm{i}},-{\mu }_{\mathrm{k}}g$for$a,anddfors$.

$(0{)}^2-{\left(v_0\right)}^2=(2)\left(-{\mu }_{\mathrm{k}}g\right)d$

$d=\frac{v_0^2}{2{\mu }_{\mathrm{k}}g}$

The distance travelled by an object has nothing to do with its mass. As a result, the distance travelled remains constant even if the object's mass is doubled.

Explanation (b)

In this situation, the object's initial velocity is doubled while its mass remains unchanged. Consider the object's distance travelled in this scenario to be$d^{\text{'}}$

Putting 0 for $v_{\mathrm{f}},2v_0$for $v_{\mathrm{i}},-{\mu }_{\mathrm{k}}g$for $a,andd\text{'}fors$in the equation $v_{\mathrm{f}}^2-v_{\mathrm{i}}^2=2a$.

$(0{)}^2-{\left(2v_0\right)}^2=(2)\left(-{\mu }_{\mathrm{k}}g\right)d^{\text{'}}$

Equation rearrange for $d^{\text{'}}$.

$d^{\text{'}}=\frac{4v_0^2}{2{\mu }_{\mathrm{k}}g}$

$=4\left(\frac{v_0^2}{2{\mu }_{\mathrm{k}}g}\right)$

$=4d$

As a result, if the original velocity is double that of the initial number, the object has travelled four times as far.