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Q25 PE

Expert-verified
Found in: Page 261

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# (a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h? (b) If, in actuality, a 750-kg car with an initial speed of 110 km/h is observed to coast up a hill to a height 22.0 m above its starting point, how much thermal energy was generated by friction? (c) What is the average force of friction if the hill has a slope 2.5° above the horizontal?

(a) The car can go up to the height of $47.63\text{m}$.

(b) The thermal energy produced due to heat is $188415.74\text{J}$.

(c) The average force of friction is $373.57\text{N}$.

See the step by step solution

## Conservation of energy

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

$\Delta \text{KE}={W}_{f}-\Delta \text{PE}\phantom{\rule{0ex}{0ex}}\frac{1}{2}m{v}_{f}^{2}-\frac{1}{2}m{v}_{i}^{2}={W}_{f}-\left(mg{h}_{f}-mg{h}_{i}\right)\phantom{\rule{0ex}{0ex}}\frac{1}{2}m{v}_{f}^{2}-\frac{1}{2}m{v}_{i}^{2}={W}_{f}-mg{h}_{f}+mg{h}_{i}$ (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).

## Maximum height attained by the car

(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),

$\left[\frac{1}{2}×\left(750\text{kg}\right)×{\left(0\right)}^{2}-\phantom{\rule{0ex}{0ex}}\frac{1}{2}×\left(750\text{kg}\right)×{\left(110\text{km}/\text{h}\right)}^{2}\right]=\left[0-\left(750\text{kg}\right)×\left(9.8\text{m}/{\text{s}}^{2}\right)×{h}_{f}\phantom{\rule{0ex}{0ex}}+\left(750\text{kg}\right)×\left(9.8\text{m}/{\text{s}}^{2}\right)×\left(0\right)\right]\phantom{\rule{0ex}{0ex}}-\frac{1}{2}×\left(750\text{kg}\right)×{\left(30.56\text{m}/\text{s}\right)}^{2}=-\left(750\text{kg}\right)×\left(9.8\text{m}/{\text{s}}^{2}\right)×{h}_{f}\phantom{\rule{0ex}{0ex}}{h}_{f}=\frac{-\frac{1}{2}×\left(750\text{kg}\right)×{\left(30.56\text{m}/\text{s}\right)}^{2}}{-\left(750\text{kg}\right)×\left(9.8\text{m}/{\text{s}}^{2}\right)}\phantom{\rule{0ex}{0ex}}{h}_{f}=47.63\text{m}$

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

## The heat energy generated

(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,

$\left[\frac{1}{2}×\left(750\text{kg}\right)×{\left(0\right)}^{2}-\phantom{\rule{0ex}{0ex}}\frac{1}{2}×\left(750\text{kg}\right)×{\left(110\text{km}/\text{h}\right)}^{2}\right]=\left[{W}_{f}-\left(750\text{kg}\right)×\left(9.8\text{m}/{\text{s}}^{2}\right)×\left(22.0\text{m}\right)\phantom{\rule{0ex}{0ex}}+\left(750\text{kg}\right)×\left(9.8\text{m}/{\text{s}}^{2}\right)×\left(0\right)\right]\phantom{\rule{0ex}{0ex}}-\frac{1}{2}×\left(750\text{kg}\right)×{\left(30.56\text{m}/\text{s}\right)}^{2}={W}_{f}-\left(750\text{kg}\right)×\left(9.8\text{m}/{\text{s}}^{2}\right)×\left(22.0\text{m}\right)\phantom{\rule{0ex}{0ex}}{W}_{f}=\left[\left(750\text{kg}\right)×\left(9.8\text{m}/{\text{s}}^{2}\right)×\left(22.0\text{m}\right)-\phantom{\rule{0ex}{0ex}}\frac{1}{2}×\left(750\text{kg}\right)×{\left(30.56\text{m}/\text{s}\right)}^{2}\right]\phantom{\rule{0ex}{0ex}}=-188415.74\text{J}$

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 average frictional force

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,

$d=\frac{h}{\mathrm{sin}\left(2.5°\right)}$

Putting all known values,

$d=\frac{\left(22.0\text{m}\right)}{\mathrm{sin}\left(2.5°\right)}\phantom{\rule{0ex}{0ex}}=504.36\text{m}$

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

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

Putting all known values,

$F=-\frac{\left(-188415.74\text{J}\right)}{\left(504.36\text{m}\right)}\phantom{\rule{0ex}{0ex}}=373.57\text{N}$

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