Chapter 5, Problem 10WUE

A puck of mass 0.170 kg slides across ice in the positive x-direction with a kinetic friction coefficient between the ice and puck of 0.150. If the puck is moving at an initial speed of 12.0 m/s, (a) what is the force of kinetic friction? (b) What is the acceleration of the puck? (c) How long does it take for the puck to come to rest? (d) What distance does the puck travel during that time? (e) What total work does friction do on the puck? (f) What average power does friction generate in the puck during that time? (g) What instantaneous power does friction generate in the puck when the velocity is 6.00 m/s? (See Sections 2.5, 4.6, 5.1, and 5.6.)

To determine

What is the force of kinetic friction.

Answer to Problem 10WUE

Solution: The force of kinetic friction is $-250\text{N}$.

Explanation of Solution

Given Info: The mass of the puck is $0.170\text{kg}$, the kinetic friction coefficient between the ice and the puck is $0.150$, the initial speed of the puck is $12.0\text{m/s}$.

The expression used for the force of kinetic friction.

$f_k=-{\mu }_{k}n$

• ${\mu }_k$ is the kinetic friction coefficient
• n is the normal force
• $f_k$ is the force of kinetic friction

Thus,

$f_k=-{\mu }_{k}mg$

• m is the mass of the puck
• g is acceleration due to gravity

Substitute $0.150$ for ${\mu }_k$, $0.170\text{kg}$ for m and $9.8{\text{m/s}}^{\text{2}}$ for g to find the force of kinetic friction.

$\begin{array}{c}{f}_k=-\left(0.150\right)\left(0.170\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\\ =-250\text{N}\end{array}$

Thus, the force of kinetic friction is $-250\text{N}$.

Conclusion:

The force of kinetic friction is $-250\text{N}$.

To determine

The acceleration of the puck.

Answer to Problem 10WUE

Solution: The acceleration of the puck is $-1.47{\text{m/s}}^{\text{2}}$.

Explanation of Solution

Given Info: The mass of the puck is $0.170\text{kg}$, the kinetic friction coefficient between the ice and the puck is $0.150$, the initial speed of the puck is $12.0\text{m/s}$, The force of kinetic friction is $-250\text{N}$.

The expression used for Newton’s second law of motion.

${\displaystyle \sum F_x}=f_k=ma$

• ${\displaystyle \sum F_x}$ is the total force in the horizontal direction
• a is the acceleration of the puck

Thus,

$a=\frac{f_k}{m}$

Substitute $-250\text{N}$ for $f_k$ and $0.170\text{kg}$ for m to find the acceleration of the puck.

$\begin{array}{c}a=\frac{-250\text{N}}{0.170\text{kg}}\\ =-1.47{\text{m/s}}^{\text{2}}\end{array}$

Thus, the acceleration of the puck is $-1.47{\text{m/s}}^{\text{2}}$.

Conclusion:

The acceleration of the puck is $-1.47{\text{m/s}}^{\text{2}}$.

To determine

The time taken for a puck to come to rest.

Answer to Problem 10WUE

Solution: The time taken for a puck to come to rest is $8.16\text{s}$.

Explanation of Solution

Given Info: The mass of the puck is $0.170\text{kg}$, the kinetic friction coefficient between the ice and the puck is $0.150$, the initial speed of the puck is $12.0\text{m/s}$, The force of kinetic friction is $-250\text{N}$, The acceleration of the puck is $-1.47{\text{m/s}}^{\text{2}}$.

The expression used for velocity of motion.

$v=v_0+at$

• $v_0$ is the initial velocity
• t is the time taken for a puck to come to rest
• v is the final velocity of the puck

Thus,

$t=\frac{v-v_0}{a}$

Substitute $0\text{m/s}$ for $v$, $12.0\text{m/s}$ for $v_0$ and $-1.47{\text{m/s}}^{\text{2}}$ for a to find time taken for a puck to come to rest.

$\begin{array}{c}t=\frac{0\text{m/s}-12.0\text{m/s}}{-1.47{\text{m/s}}^{\text{2}}}\\ =8.16\text{s}\end{array}$

Thus, the time taken for a puck to come to rest is $8.16\text{s}$.

Conclusion:

The time taken for a puck to come to rest is $8.16\text{s}$.

To determine

The distance the puck travelled during the time.

Answer to Problem 10WUE

Solution: The distance the puck travelled during the time is $49.0\text{m}$.

Explanation of Solution

Given Info: The mass of the puck is $0.170\text{kg}$, the kinetic friction coefficient between the ice and the puck is $0.150$, the initial speed of the puck is $12.0\text{m/s}$, The force of kinetic friction is $-250\text{N}$, The time taken for a puck to come to rest is $8.16\text{s}$.

The expression used for distance the puck travelled during the time.

$\Delta x=v_{0}t+\frac{1}{2}a{t}^2$

• $\Delta x$ is the distance the puck travelled during the time

Substitute $12.0\text{m/s}$ for $v_0$, $8.16\text{s}$ for t and $-1.47{\text{m/s}}^{\text{2}}$ for a to find the distance the puck travelled during the time.

$\begin{array}{c}\Delta x=\left(12.0\text{m/s}\right)\left(8.16\text{s}\right)+\frac{1}{2}\left(-1.47{\text{m/s}}^{\text{2}}\right){\left(8.16\text{s}\right)}^2\\ =49.0\text{m}\end{array}$

Thus, the distance the puck travelled during the time is $49.0\text{m}$.

Conclusion:

The distance the puck travelled during the time is $49.0\text{m}$.

To determine

The total work done by the friction on the puck.

Answer to Problem 10WUE

Solution: The total work done by the friction on the puck is $-12.2\text{J}$.

Explanation of Solution

Given Info: The mass of the puck is $0.170\text{kg}$, the kinetic friction coefficient between the ice and the puck is $0.150$, the initial speed of the puck is $12.0\text{m/s}$, The force of kinetic friction is $-250\text{N}$, The time taken for a puck to come to rest is $8.16\text{s}$, The distance the puck travelled during the time is $49.0\text{m}$.

The expression used for total work done by the friction on the puck.

$W_f=\left(f_k\cos 180°\right)\Delta x$

• $\Delta x$ is the distance the puck travelled during the time
• $W_f$ is the total work done by the friction on the puck

Substitute $-250\text{N}$ for $f_k$, and $49.0\text{m}$ for $\Delta x$ to find the total work done by the friction on the puck.

$\begin{array}{c}{W}_f=\left(\left(-250\text{N}\right)\cos 180°\right)49.0\text{m}\\ =-12.2\text{J}\end{array}$

Thus, the total work done by the friction on the puck is $-12.2\text{J}$.

Conclusion:

The total work done by the friction on the puck is $-12.2\text{J}$.

To determine

The average power generated by the friction during the time.

Answer to Problem 10WUE

Solution: The average power generated by the friction during the time is $-1.50\text{W}$.

Explanation of Solution

Given Info: The mass of the puck is $0.170\text{kg}$, the kinetic friction coefficient between the ice and the puck is $0.150$, the initial speed of the puck is $12.0\text{m/s}$, The force of kinetic friction is $-250\text{N}$, The time taken for a puck to come to rest is $8.16\text{s}$, The distance the puck travelled during the time is $49.0\text{m}$.

The expression used for average power generated by the friction.

$\begin{array}{c}\overline{P}=F\overline{v}\\ =f_k\left(\frac{v+v_0}{2}\right)\end{array}$

• $\overline{P}$ is the average power
• $F$ is the force
• $\overline{v}$ is the average velocity

Substitute $-250\text{N}$ for $f_k$, $0\text{m/s}$ for v and $12\text{m/s}$ for $v_0$ to find the for average power generated by the friction.

$\begin{array}{c}\overline{P}=-250\text{N}\left(\frac{0+12\text{m/s}}{2}\right)\\ =-1.50\text{W}\end{array}$

Thus, the average power generated by the friction during the time is $-1.50\text{W}$.

Conclusion:

The average power generated by the friction during the time is $-1.50\text{W}$.

To determine

The instantaneous power the friction generate in the puck when the velocity is $6.0\text{m/s}$.

Answer to Problem 10WUE

Solution: The instantaneous power the friction generate in the puck is $-1.50\text{W}$.

Explanation of Solution

Given Info: The mass of the puck is $0.170\text{kg}$, the kinetic friction coefficient between the ice and the puck is $0.150$, the initial speed of the puck is $12.0\text{m/s}$, The force of kinetic friction is $-250\text{N}$, The time taken for a puck to come to rest is $8.16\text{s}$, The distance the puck travelled during the time is $49.0\text{m}$.

The expression used for the instantaneous power the friction generate in the puck.

$\begin{array}{c}P=Fv\\ =f_{k}v\end{array}$

• P is the instantaneous power
• v is the velocity of the puck

Substitute $-250\text{N}$ for $f_k$ and $6.0\text{m/s}$ for $v$ to find the instantaneous power the friction generate in the puck.

$\begin{array}{c}P=\left(-250\text{N}\right)\left(6.0\text{m/s}\right)\\ =-1.50\text{m}\end{array}$

Thus, the instantaneous power the friction generate in the puck is $-1.50\text{W}$.

Conclusion:

The instantaneous power the friction generate in the puck is $-1.50\text{W}$