Chapter 7, Problem 63P

(a)

To determine

The work done by the applied force.

Explanation of Solution

Given:

The mass of sled is $8.0\text{kg}$ .

The coefficient of friction between sled and road is $0.40$ .

The distance traveled by sled is $3.0\text{m}$ .

The force applied to the sled is $40\text{N}$ at an angle of $30°$ with the horizontal.

Formula used:

Write the expression for work done by external force.

  $W=Fs\cos \theta$ ........(1)

Here, $F$ is the external force, $s$ is the displacement, $\theta$ is the angle between displacement and force and $W$ is the work done.

Calculation:

Substitute $40\text{N}$ for $F$ , $3.0\text{m}$ for $s$ and $30°$ for $\theta$ in equation (1).

  $\begin{array}{c}W=\left(40\text{N}\right)\left(3.0\text{m}\right)\cos 30°\\ =103.9\text{J}\end{array}$

Conclusion:

Thus, the work done by external force is $103.9\text{J}$ .

(b)

To determine

Theenergy dissipated by friction.

Explanation of Solution

Given:

The mass of sled is $8.0\text{kg}$ .

The coefficient of friction between sled and road is $0.40$ .

The distance traveled by sled is $3.0\text{m}$ .

The force applied to the sled is $40\text{N}$ at an angle of $30°$ with the horizontal.

Formula used:

Write the expression for friction force.

  $f=\mu F_n$

Here, $F_n$ is the normal reaction, $f$ is the friction force and $\mu$ is the coefficient of kinetic friction.

Write the expression for thermal energy.

  $\Delta E_{\text{thermal}}=fs$

Here, $\Delta E_{\text{thermal}}$ is the thermal energy produced and $s$ is the displacement covered by sled.

Substitute $\mu F_n$ for $f$ in above expression.

  $\Delta E_{\text{thermal}}=\mu F_{n}s$ ........(2)

The free body diagram of sled is given below.

  

Write the expression for resultant force in vertical direction.

  $F_n+F\sin \theta =F_g$

Here, $F$ is the external force, $\theta$ is the angle of force with the horizontal and $F_g$ is the weight of sled.

Substitute $mg$ for $F_g$ in above expression and rearrange in terms of $F_n$ .

  $F_n=mg-F\sin \theta$

Substitute $mg-F\sin \theta$ for $F_n$ in equation (2).

  $\Delta E_{\text{thermal}}=\mu s\left(mg-F\sin \theta \right)$ ........(3)

Calculation:

Substitute $40\text{N}$ for $F$ , $3.0\text{m}$ for $s$ , $8.0\text{kg}$ for $m$ , $0.40$ for $\mu$ , $9.81\text{m}/{\text{s}}^2$ for $g$ and $30°$ for $\theta$ in equation (3).

  $\begin{array}{c}\Delta E_{\text{thermal}}=\left(0.40\right)\left(3.0\text{m}\right)\left[\left(8.0\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)-\left(40\text{N}\right)\sin 30°\right]\\ =70.18\text{J}\end{array}$

Conclusion:

Thus, the energy dissipated by frictionis $70.18\text{J}$ .

(c)

To determine

The change in kinetic energy of the sled.

Explanation of Solution

Given:

The mass of sled is $8.0\text{kg}$ .

The coefficient of friction between sled and road is $0.40$ .

The distance traveled by sled is $3.0\text{m}$ .

The force applied to the sled is $40\text{N}$ at an angle of $30°$ with the horizontal.

Formula used:

Write the expression for work done by external force.

  $W=Fs\cos \theta$

Here, $F$ is the external force, $s$ is the displacement, $\theta$ is the angle between displacement and force and $W$ is the work done.

Write the expression for friction force.

  $f=\mu F_n$

Here, $F_n$ is the normal reaction, $f$ is the friction force and $\mu$ is the coefficient of kinetic friction.

Write the expression for thermal energy.

  $\Delta E_{\text{thermal}}=fs$

Here, $\Delta E_{\text{thermal}}$ is the thermal energy produced and $s$ is the displacement covered by sled.

Substitute $\mu F_n$ for $f$ in above expression.

  $\Delta E_{\text{thermal}}=\mu F_{n}s$

The free body diagram of sled is given below.

  

Write the expression for resultant force in vertical direction.

  $F_n+F\sin \theta =F_g$

Here, $F$ is the external force, $\theta$ is the angle of force with the horizontal and $F_g$ is the weight of sled.

Substitute $mg$ for $F_g$ in above expression and rearrange in terms of $F_n$ .

  $F_n=mg-F\sin \theta$

Substitute $mg-F\sin \theta$ for $F_n$ in equation (2).

  $\Delta E_{\text{thermal}}=\mu s\left(mg-F\sin \theta \right)$ ........(3)

Total energy of sled is conserved at all points. Work done by external force is equal to the sum of change in gravitational potential energy, kinetic energy and thermal energy.

Write the expression of work done by external force.

  $W=\Delta U+\Delta K+\Delta E_{\text{thermal}}$

Here, $\Delta U$ is the potential energy stored in the body and $\Delta K$ is the change in kinetic energy.

The height of sled is constant all the time; so, change in potential energy is zero.

Substitute $0$ for $\Delta U$ in above equation and rearrange in terms of $\Delta K$ .

  $\Delta K=W-\Delta E_{\text{thermal}}$ ........(4)

Calculation:

Substitute $40\text{N}$ for $F$ , $3.0\text{m}$ for $s$ and $30°$ for $\theta$ in equation (1).

  $\begin{array}{c}W=\left(40\text{N}\right)\left(3.0\text{m}\right)\cos 30°\\ =103.9\text{J}\end{array}$

Substitute $40\text{N}$ for $F$ , $3.0\text{m}$ for $s$ , $8.0\text{kg}$ for $m$ , $0.40$ for $\mu$ , $9.81\text{m}/{\text{s}}^2$ for $g$ and $30°$ for $\theta$ in equation (3).

  $\begin{array}{c}\Delta E_{\text{thermal}}=\left(0.40\right)\left(3.0\text{m}\right)\left[\left(8.0\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)-\left(40\text{N}\right)\sin 30°\right]\\ =70.18\text{J}\end{array}$

Substitute $103.9\text{J}$ for $W$ and $70.18\text{J}$ for $\Delta E_{\text{thermal}}$ in equation (4).

  $\begin{array}{c}\Delta K=103.9\text{J}-70.18\text{J}\\ =33.72\text{J}\end{array}$

Conclusion:

Thus, the change in kinetic energy of the sled is $33.72\text{J}$ .

(d)

To determine

The speed of sled after it has traveled $3.0\text{m}$ distance.

Explanation of Solution

Given:

The mass of sled is $8.0\text{kg}$ .

The coefficient of friction between sled and road is $0.40$ .

The distance traveled by sled is $3.0\text{m}$ .

The force applied to the sled is $40\text{N}$ at an angle of $30°$ with the horizontal.

Formula used:

Write the expression for work done by external force.

  $W=Fs\cos \theta$

Here, $F$ is the external force, $s$ is the displacement, $\theta$ is the angle between displacement and force and $W$ is the work done.

Write the expression for friction force.

  $f=\mu F_n$

Here, $F_n$ is the normal reaction, $f$ is the friction force and $\mu$ is the coefficient of kinetic friction.

Write the expression for thermal energy.

  $\Delta E_{\text{thermal}}=fs$

Here, $\Delta E_{\text{thermal}}$ is the thermal energy produced and $s$ is the displacement covered by sled.

Substitute $\mu F_n$ for $f$ in above expression.

  $\Delta E_{\text{thermal}}=\mu F_{n}s$

The free body diagram of sled is given below.

  

Write the expression for resultant force in vertical direction.

  $F_n+F\sin \theta =F_g$

Here, $F$ is the external force, $\theta$ is the angle of force with the horizontal and $F_g$ is the weight of sled.

Substitute $mg$ for $F_g$ in above expression and rearrange in terms of $F_n$ .

  $F_n=mg-F\sin \theta$

Substitute $mg-F\sin \theta$ for $F_n$ in equation (2).

  $\Delta E_{\text{thermal}}=\mu s\left(mg-F\sin \theta \right)$ ........(3)

Total energy of sled is conserved at all points. Work done by external force is equal to the sum of change in gravitational potential energy, kinetic energy and thermal energy.

Write the expression of work done by external force.

  $W=\Delta U+\Delta K+\Delta E_{\text{thermal}}$

Here, $\Delta U$ is the potential energy stored in the body and $\Delta K$ is the change in kinetic energy.

The height of car is constant all the time; so, change in potential energy is zero.

Substitute $0$ for $\Delta U$ in above equation and rearrange in terms of $\Delta K$ .

  $\Delta K=W-\Delta E_{\text{thermal}}$ ........(4)

Write the expression for change in kinetic energy.

  $\Delta K=\frac{1}{2}m\left(v_f^2-v_i^2\right)$

Rearrange the above expression in terms of $v_f$ .

  $v_f=\sqrt{\frac{2\Delta K}{m}+v_i^2}$ ........(5)

Here, $v_i$ is the initial velocity and $v_f$ is the final velocity of sled.

Calculation:

Substitute $40\text{N}$ for $F$ , $3.0\text{m}$ for $s$ and $30°$ for $\theta$ in equation (1).

  $\begin{array}{c}W=\left(40\text{N}\right)\left(3.0\text{m}\right)\cos 30°\\ =103.9\text{J}\end{array}$

Substitute $40\text{N}$ for $F$ , $3.0\text{m}$ for $s$ , $8.0\text{kg}$ for $m$ , $0.40$ for $\mu$ , $9.81\text{m}/{\text{s}}^2$ for $g$ and $30°$ for $\theta$ in equation (3).

  $\begin{array}{c}\Delta E_{\text{thermal}}=\left(0.40\right)\left(3.0\text{m}\right)\left[\left(8.0\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)-\left(40\text{N}\right)\sin 30°\right]\\ =70.18\text{J}\end{array}$

Substitute $103.9\text{J}$ for $W$ and $70.18\text{J}$ for $\Delta E_{\text{thermal}}$ in equation (4).

  $\begin{array}{c}\Delta K=103.9\text{J}-70.18\text{J}\\ =33.72\text{J}\end{array}$

Substitute $0$ for $v_i$ , $8.0\text{kg}$ for $m$ and $33.72\text{J}$ for $\Delta K$ in equation (5).

  $\begin{array}{c}{v}_f=\sqrt{\frac{2\left(33.72\text{J}\right)}{8.0\text{kg}}+0}\\ =2.90\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the speed of sled after it has traveled $3.0\text{m}$ distance is $2.90\text{m}/\text{s}$ .