Chapter 12, Problem 3Q

(a) For the car going up the incline, what percentage of the work done by the suspended weight was lost to friction? (b) For the car moving down the incline, what percentage of the work done by gravity was lost to friction? (Show your calculations.)

To determine

The percentage of work done by the suspended weight that is lost to friction when a car is moving up an incline with constant speed.

Answer to Problem 3Q

The percentage loss of energy by the suspended weight is $\left[1-\frac{2d}{l}\left(1-\frac{M}{2m}\right)\right]\%$ for $\theta =30°$ and $\left[1-\frac{\sqrt{2}d}{l}\left(1-\frac{M}{\sqrt{2}m}\right)\right]\%$ for $\theta =45°$.

Explanation of Solution

Draw a diagram to show the different components of forces acting on the car.

Write the expression for the net horizontal force acting on the car

    $F_n=F-Mg\sin \theta -f$        (I)

Here, $F_n$ is the net force, $F$ is the force acted upon the car by the pull of the mass $m$, $\theta$ is the angle of inclination and $f$ is the frictional force.

Write the expression for the force acted upon the car by the pull of the mass $m$.

    $F=mg$

Here, $g$ is the acceleration due to gravity.

Write the expression for the work done by the frictional force.

    $W_f=fd$        (II)

Here, $W_f$ is the work done and $d$ is the distance travelled by the car along the inclined plane.

Write the expression for the percentage loss of energy.

    $P=[1-r]\%$        (III)

Here, $P$  is the percentage of the lost energy and $r$ is the ratio of the work done by the friction force to the work done by the suspended weight.

Write the expression for the work done by the suspended weight.

    $W=mgh$        (IV)

Here, $W$ is the work done, $m$ is the mass of the suspended weight, $g$ is the acceleration due to gravity and $h$ is the vertical height traveled by the weight.

Conclusion:

Substitute $0$ for $F_n$ in expression (I) as the car is moving upward at a constant speed and rearrange it.

    $f=F-Mg\text{sin}\theta$

Substitute $mg$ for $F$ in the above expression.

    $f=mg-Mg\text{sin}\theta$

Substitute $mg-Mg\text{sin}\theta$ for $f$ in the expression (II).

    $W_f=g\left(m-M\text{sin}\theta \right)d$        (V)

For $\theta =30°$,

Substitute $W_{30}$  for $W$ and $l/2$  for $h$  in expression (IV).

    $W_{30}=\frac{1}{2}mgl$        (VI)

Substitute $30°$ for $\theta$ in expression (V).

    $W_f=g\left(m-\frac{M}{2}\right)d$        (VII)

Divide expression (VII) by (VI).

    $\begin{array}{c}{r}_{30}=\frac{W_f}{W_{30}}\\ =\frac{2d}{l}\left(1-\frac{M}{2m}\right)\end{array}$

Here, $r_{30}$ is the ratio of the work done by the friction force to the work done by the suspended weight.

Substitute $P_{30}$ for $P$ and $\frac{2d}{l}\left(1-\frac{M}{2m}\right)$ for $r$ in expression (III).

    $P_{30}=\left[1-\frac{2d}{l}\left(1-\frac{M}{2m}\right)\right]\%$

For $\theta =45°$,

Substitute $W_{45}$  for $W$ and $l/\sqrt{2}$  for $h$  in expression (IV).

    $W_{45}=\frac{1}{\sqrt{2}}mgl$        (VIII)

Substitute $45°$ for $\theta$ in expression (V).

    $W_f=g\left(m-\frac{M}{\sqrt{2}}\right)d$        (IX)

Divide expression (IX) by (VIII).

    $\begin{array}{c}{r}_{45}=\frac{W_f}{W_{45}}\\ =\frac{\sqrt{2}d}{l}\left(1-\frac{M}{\sqrt{2}m}\right)\end{array}$

Here, $r_{45}$ is the ratio of the work done by the friction force to the work done by the suspended weight.

Substitute $P_{45}$ for $P$ and $\frac{\sqrt{2}d}{l}\left(1-\frac{M}{\sqrt{2}m}\right)$ for $r$ in expression (III).

    $P_{45}=\left[1-\frac{\sqrt{2}d}{l}\left(1-\frac{M}{\sqrt{2}m}\right)\right]\%$

Thus, the percentage loss of energy by the suspended weight is $\left[1-\frac{2d}{l}\left(1-\frac{M}{2m}\right)\right]\%$ for $\theta =30°$ and $\left[1-\frac{\sqrt{2}d}{l}\left(1-\frac{M}{\sqrt{2}m}\right)\right]\%$ for $\theta =45°$.

To determine

The percentage of work done by gravity that is lost to friction when a car is moving down an incline with constant speed.

Answer to Problem 3Q

The percentage loss of energy by gravity is $\left[1-\frac{2d}{l}\left(\frac{M}{2m}-1\right)\right]\%$ for $\theta =30°$ and $\left[1-\frac{\sqrt{2}d}{l}\left(\frac{M}{\sqrt{2}m}-1\right)\right]\%$ for $\theta =45°$.

Explanation of Solution

Draw a diagram to show the different components of forces acting on the car.

Write the expression for the net horizontal force acting on the car

    $F_n=F-Mg\text{sin}\theta +f$        (X)

Here, $F_n$ is the net force, $F$ is the force acted upon the car by the pull of the mass $m$, $\theta$ is the angle of inclination and $f$ is the frictional force.

Write the expression for the force acted upon the car by the pull of the mass $m$.

    $F=mg$

Here, $g$ is the acceleration due to gravity.

Write the expression for the work done by the frictional force.

    $W_f=fd$        (XI)

Here, $W_f$ is the work done and $d$ is the distance travelled by the car along the inclined plane.

Write the expression for the percentage loss of energy.

    $P=[1-r]\%$        (XII)

Here, $P$  is the percentage of the lost energy and $r$ is the ratio of the work done by the friction force to the work done by gravity.

Write the expression for the work done by the suspended weight.

    $W=mgh$        (XIII)

Here, $W$ is the work done, $m$ is the mass of the suspended weight, $g$ is the acceleration due to gravity and $h$ is the vertical height traveled by the weight.

Conclusion:

Substitute $0$ for $F_n$ in expression (IX) as the car is moving downward at a constant speed and rearrange it.

    $f=Mg\text{sin}\theta -F$        (XIV)

Substitute $mg$ for $F$ in the above expression.

    $f=Mg\text{sin}\theta -mg$

Substitute $Mg\text{sin}\theta -mg$ for $f$ in expression (X).

    $W_f=g\left(M\text{sin}\theta -m\right)d$        (XV)

For $\theta =30°$,

Substitute $W_{30}$  for $W$ and $l/2$  for $h$  in expression (XIII).

    $W_{30}=\frac{1}{2}mgl$        (XVI)

Here, $W_{30}$ is the work done by gravity when the angle is $30°$.

Substitute $30°$ for $\theta$ in expression (XV).

    $W_f=g\left(\frac{M}{2}-m\right)d$        (XVII)

Divide expression (XVII) by (XVI).

    $\begin{array}{c}{r}_{30}=\frac{W_f}{W_{30}}\\ =\frac{2d}{l}\left(\frac{M}{2m}-1\right)\end{array}$

Here, $r_{30}$ is the ratio of the work done by the friction force to the work done by gravity.

Substitute $P_{30}$ for $P$ and $\frac{2d}{l}\left(1-\frac{M}{2m}\right)$ for $r$ in expression (XII).

    $P_{30}=\left[1-\frac{2d}{l}\left(\frac{M}{2m}-1\right)\right]\%$

Here, $P_{30}$ is the percentage loss of energy.

For $\theta =45°$,

Substitute $W_{30}$  for $W$ and $l/\sqrt{2}$  for $h$  in expression (XIII).

    $W_{45}=\frac{1}{\sqrt{2}}mgl$        (XVIII)

Here, $W_{45}$ is the work done by gravity when the angle is $45°$.

Substitute $45°$ for $\theta$ in expression (XV).

    $W_f=g\left(\frac{M}{\sqrt{2}}-m\right)d$        (XIX)

Divide expression (XIX) by (XVIII).

    $\begin{array}{c}{r}_{45}=\frac{W_f}{W_{45}}\\ =\frac{\sqrt{2}d}{l}\left(\frac{M}{\sqrt{2}m}-1\right)\end{array}$

Here, $r_{45}$ is the ratio of the work done by the friction force to the work done by gravity.

Substitute $P_{45}$ for $P$ and $\frac{\sqrt{2}d}{l}\left(1-\frac{M}{\sqrt{2}m}\right)$ for $r$ in expression (XII).

    $P_{45}=\left[1-\frac{\sqrt{2}d}{l}\left(\frac{M}{\sqrt{2}m}-1\right)\right]\%$

Here, $P_{45}$ is the percentage loss of energy.

Thus, the percentage loss of energy by gravity is $\left[1-\frac{2d}{l}\left(\frac{M}{2m}-1\right)\right]\%$ for $\theta =30°$ and $\left[1-\frac{\sqrt{2}d}{l}\left(\frac{M}{\sqrt{2}m}-1\right)\right]\%$ for $\theta =45°$.