91Ó°ÊÓ

Conservation of energy: When both conservative and nonconservative force acts on a body, the conservation of energy is given as,

$$\begin{gathered} \mathrm{Δ}\text{KE}=W_f-\mathrm{Δ}\text{PE} \\ \frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2=W_f-\left(mg{h}_f-mg{h}_i\right) \\ \frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2=W_f-mg{h}_f+mg{h}_i \end{gathered}$$ (1.1)

Here, m is the mass of the car, $v_f$is the final velocity of the car ($v_f=0$ as the car stops), $v_i$is the initial velocity of the car $\left(v_i=110\text{km}/\text{h}\right)$, $W_f$is the work done by nonconservative force or the work done by the friction, g is the acceleration due to gravity $\left(9.8\text{m}/{\text{s}}^2\right)$, $h_f$is the final height, and $h_i$is the initial height (,$h_i=0$ as the object starts from the ground).

(a)

When there is no frictional force, the work done by the frictional force will be zero i.e.,$W_f=0$.

The maximum height attained by the car can be calculated using equation (1.1).

Putting all known values in equation (1.1),

$$\begin{gathered} \left[\frac{1}{2}×\left(750\text{kg}\right)×{\left(0\right)}^2- \\ \frac{1}{2}×\left(750\text{kg}\right)×{\left(110\text{km}/\text{h}\right)}^2\right]=\left[0-\left(750\text{kg}\right)×\left(9.8\text{m}/{\text{s}}^2\right)×h_f \\ +\left(750\text{kg}\right)×\left(9.8\text{m}/{\text{s}}^2\right)×\left(0\right)\right] \\ -\frac{1}{2}×\left(750\text{kg}\right)×{\left(30.56\text{m}/\text{s}\right)}^2=-\left(750\text{kg}\right)×\left(9.8\text{m}/{\text{s}}^2\right)×h_f \\ {h}_f=\frac{-\frac{1}{2}×\left(750\text{kg}\right)×{\left(30.56\text{m}/\text{s}\right)}^2}{-\left(750\text{kg}\right)×\left(9.8\text{m}/{\text{s}}^2\right)} \\ {h}_f=47.63\text{m} \end{gathered}$$

Therefore, the car can go up to the height of $47.63\text{m}$.

(b)

When the frictional force is applicable the car can attain maximum of$h_f=22.0\text{m}$. The work done by nonconservative force or the frictional force can be calculated using equation (1.1).

Putting all known values,

$$\begin{gathered} \left[\frac{1}{2}×\left(750\text{kg}\right)×{\left(0\right)}^2- \\ \frac{1}{2}×\left(750\text{kg}\right)×{\left(110\text{km}/\text{h}\right)}^2\right]=\left[W_f-\left(750\text{kg}\right)×\left(9.8\text{m}/{\text{s}}^2\right)×\left(22.0\text{m}\right) \\ +\left(750\text{kg}\right)×\left(9.8\text{m}/{\text{s}}^2\right)×\left(0\right)\right] \\ -\frac{1}{2}×\left(750\text{kg}\right)×{\left(30.56\text{m}/\text{s}\right)}^2=W_f-\left(750\text{kg}\right)×\left(9.8\text{m}/{\text{s}}^2\right)×\left(22.0\text{m}\right) \\ {W}_f=\left[\left(750\text{kg}\right)×\left(9.8\text{m}/{\text{s}}^2\right)×\left(22.0\text{m}\right)- \\ \frac{1}{2}×\left(750\text{kg}\right)×{\left(30.56\text{m}/\text{s}\right)}^2\right] \\ =-188415.74\text{J} \end{gathered}$$

Since, the work done by the frictional force is liberated as thermal energy.

Therefore, the thermal energy produced due to heat is $188415.74\text{J}$.

The work done by the frictional force is,

$W_f=-Fd$ (1.2)

Here, F is the average frictional force and d is the distance travelled.

The distance travelled is given as,

$d=\frac{h}{\sin \left(2.5°\right)}$

Putting all known values,

$$\begin{gathered} d=\frac{\left(22.0\text{m}\right)}{\sin \left(2.5°\right)} \\ =504.36\text{m} \end{gathered}$$

Rearranging equation (1.2) in order to get average frictional force.

$F=-\frac{W}{d}$

Putting all known values,

$$\begin{gathered} F=-\frac{\left(-188415.74\text{J}\right)}{\left(504.36\text{m}\right)} \\ =373.57\text{N} \end{gathered}$$

Therefore, the required average frictional force is$373.57\text{N}$