Question

A crate push along the floor with velocity v slides a distance d after the pushing force is removed. If the mass of the crate is doubled but the initial velocity is not changed, what distance does the crate slide before stopping? Explain. If the initial velocity of the crate is double to 2v but the mass is not changed, what distance does the crate slide before stoppingexplain

Answer 1

Concepts and reason

The concepts used to solve the problem are Newton’s law of motion and friction.

First calculate the final velocity of the object in terms of the acceleration and after that calculate the frictional force and compare the acceleration result.

From the final velocity expression it depends on the acceleration of the object and the frictional force depends on the mass and coefficient of friction.

Fundamentals

From Newton’s second law the expression of the force is equal to,

$F = ma$

Here, $F$ is the force, $m$ is the mass and $a$ is the acceleration of the object.

From linear kinematics, the expression of the final velocity in terms of the acceleration is equal to,

${v^2} = {u^2} + 2as$

Here, $v$ is the final velocity, $u$ is the initial velocity and $s$ is the distance covered by the object.

The expression of the friction force is equal to,

${F_r} = \mu N$

Here, ${F_r}$ is the frictional force, $\mu$ is the coefficient of the friction and $N$ is the normal force.

Calculate the acceleration of object.

The expression of the final velocity is equal to,

${v^2} = {u^2} + 2as$

Substitute $0$ for $u$ in the above expression of the final velocity and rearrange it.

$a = \frac{{{v^2}}}{{2s}}$

The expression of the force in terms of the acceleration of the object is as follows,

$F = ma$

Apply the equilibrium condition of force in the horizontal direction.

$\mu N = ma$

Substitute $mg$ for $N$ in the above expression of the force,

$\begin{array}{c}\\\mu mg = ma\\\\a = \mu g\\\end{array}$

The crate moves with a constant velocity $v$ .

Substitute $\frac{{{v^2}}}{{2s}}$ for $a$ in the above equation.

$\begin{array}{c}\\\frac{{{v^2}}}{{2s}} = \mu g\\\\s = \frac{{{v^2}}}{{2\mu g}}\\\end{array}$

Since, distance does not depends on the mass of the object, the distance covered by the crate will be the same irrespective of the mass of the object.

The newton’s equation of motion relating the initial and final velocity, distance and acceleration is,

${v^2} = {u^2} + 2as$

Rearrange the above equation.

$s = \frac{{{v^2} - {u^2}}}{{2a}}$

Substitute $0$ for $u$ in the above expression of the distance,

$s = \frac{{{v^2}}}{{2a}}$

Substitute $2v$ for $v$ and ${s_1}$ for $s$ in the above expression of the distance,

${s_1} = \frac{{4{v^2}}}{{2a}}$

Divide the distance $s$ and ${s_1}$ .

$\begin{array}{c}\\\frac{s}{{{s_1}}} = \frac{{\left( {\frac{{{v^2}}}{{2a}}} \right)}}{{\left( {\frac{{4{v^2}}}{{2a}}} \right)}}\\\\ = \frac{1}{4}\\\end{array}$

Rearrange the above expression of the distance in terms of the ${s_1}$ .

${s_1} = 4s$

Ans:

The distance covered by the crate when the mass id doubled will be $d$ .

If the velocity is doubled, then distance is equal to 4 times the initial distance.