Question

In a truck-loading station at a post office, a small 0.200-kg package is released from rest at point A on a track that is one-quarter of a circle with radius 1.60 m (the figure (Figure 1) ). The size of the package is much less than 1.60 m, so the package can be treated as a particle. It slides down the track and reaches point B with a speed of 4.00m/s . From point B, it slides on a level surface a distance of 3.00 m to point C, where it comes to rest.

In a truck-loading station at a post office, a small 0.200-kg package is released from rest at point A on a track that is one-quarter of a circle with radius 1.60 m (the figure (Figure 1) ). The size of the package is much less than 1.60 m, so the package can be treated as a particle. It slides down the track and reaches point B with a speed of 4.00m/s . From point B, it slides on a level surface a distance of 3.00 m to point C, where it comes to rest. 1-What is the coefficient of kinetic friction on the horizontal surface? 2-How much work is done on the package by friction as it slides down the circular arc from A to B?

1-What is the coefficient of kinetic friction on the horizontal surface?

2-How much work is done on the package by friction as it slides down the circular arc from A to B?

Answer 1

Concepts and reason

The concepts required to solve this problem are friction force, normal force, weight, work energy theorem and conservation of energy.

Initially, use the kinetic friction formula in the work energy theorem to solve for the coefficient of kinetic friction.

Finally, use the conservation of energy to solve for work done by friction force.

Fundamentals

The work energy theorem gives that work done by the force is equal to the change in the kinetic energy. It is expressed as follows:

$W = \Delta K$

Here, $W$ is the work done by the force and $\Delta K$ is the change in the kinetic energy.

The conservation of energy theorem states that the sum of initial potential energy, kinetic energy and work done by any force is equal to the sum of final potential energy and kinetic energy. The equation of conservation of energy is,

${U_{\rm{i}}} + {K_{\rm{i}}} + W = {U_{\rm{f}}} + {K_{\rm{f}}}$

Here, $U$ is the gravitational potential energy, $K$ is the kinetic energy, and $W$ is the work done by any force, and Subscripts ${\rm{ i and f}}$ represent initial and final state.

The gravitational potential energy is,

$U = mgy$

Here, $m$ is mass, $g$ is acceleration due to gravity, and $y$ is the vertical height of the object.

The kinetic energy is,

$K = \frac{1}{2}m{v^2}$

Here, $m$ is mass, and $v$ is the speed of the object.

The change in kinetic energy is,

$\Delta K = \frac{1}{2}m\left( {v_{\rm{i}}^2 - v_{\rm{f}}^2} \right)$

Here, $m$ is the mass, ${v_{\rm{i}}}$ is the initial speed, and ${v_{\rm{f}}}$ is the final speed.

The kinetic friction over a horizontal surface is given as,

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

Here, ${\mu _{\rm{k}}}$ is the coefficient of the kinetic friction, $m$ is the mass, and $g$ is the acceleration due to gravity.

The work done by a force is,

$W = fd$

Here, $f$ is any force, and $d$ is the distance travelled by the object.

(1)

Use the work done equation.

Substitute ${\mu _{\rm{k}}}mg$ for $f$ in the work done equation $W = fd$ to solve for the work done by the kinetic friction.

$W = {\mu _{\rm{k}}}mgd$

Use the work energy theorem to solve for the coefficient of kinetic friction.

Substitute ${\mu _{\rm{k}}}mgd$ for $W$ , and $\frac{1}{2}m\left( {v_{\rm{i}}^2 - v_{\rm{f}}^2} \right)$ for $\Delta K$ in the equation $W = \Delta K$ and solve for coefficient of kinetic friction.

$\begin{array}{c}\\{\mu _{\rm{k}}}mgd = \frac{1}{2}m\left( {v_{\rm{i}}^2 - v_{\rm{f}}^2} \right)\\\\{\mu _{\rm{k}}} = \frac{{\left( {v_{\rm{i}}^2 - v_{\rm{f}}^2} \right)}}{{2gd}}\\\end{array}$

Substitute $4.00{\rm{ m/s}}$ for ${v_{\rm{i}}}$ , $0$ for ${v_{\rm{f}}}$ , $9.8{\rm{ m/}}{{\rm{s}}^2}$ for $g$ , and $3.00{\rm{ m}}$ for $d$ in the equation ${\mu _{\rm{k}}} = \frac{{\left( {v_{\rm{i}}^2 - v_{\rm{f}}^2} \right)}}{{2gd}}$ .

$\begin{array}{c}\\{\mu _{\rm{k}}} = \frac{{\left( {{{\left( {4.00{\rm{ m/s}}} \right)}^2} - {0^2}} \right)}}{{2\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {3.0{\rm{ m}}} \right)}}\\\\ = 0.27\\\end{array}$

(2)

Use the conservation of energy equation.

Substitute $mgy$ for ${U_{\rm{i}}}$ , $0$ for ${K_{\rm{i}}}$ , 0 for ${U_{\rm{f}}}$ , and $\frac{1}{2}m{v^2}$ for ${K_{\rm{f}}}$ in the equation ${U_{\rm{i}}} + {K_{\rm{i}}} + W = {U_{\rm{f}}} + {K_{\rm{f}}}$ and solve work done $W$ .

$\begin{array}{c}\\mgy + 0 + W = 0 + \frac{1}{2}m{v^2}\\\\W = \frac{1}{2}m{v^2} - mgy\\\end{array}$

Substitute $0.200{\rm{ kg}}$ for $m$ , $4.00{\rm{ m/s}}$ for $v$ , $9.8{\rm{ m/}}{{\rm{s}}^2}$ for $g$ , and $1.60{\rm{ m}}$ for $y$ in the above equation $W = \frac{1}{2}m{v^2} - mgy$ and calculate the work done.

$\begin{array}{c}\\W = \frac{1}{2}\left( {0.200{\rm{ kg}}} \right){\left( {4.00{\rm{ m/s}}} \right)^2} - \left( {0.200{\rm{ kg}}} \right)\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {1.6{\rm{ m}}} \right)\\\\ = - 1.54{\rm{ J}}\\\end{array}$

Ans: Part 1

The coefficient of kinetic friction on the horizontal surface is $0.27$ .

Part 2

Work done on the package by friction as it slides down the circular arc from A to B is $- 1.54{\rm{ J}}$ .