Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

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Short Answer

(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 $47.63 m$ .

(b) The thermal energy produced due to heat is $188415.74 J$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Conservation of energy

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

$Δ KE = W_{f} - Δ PE \frac{1}{2} m v_{f}^{2} - \frac{1}{2} m v_{i}^{2} = W_{f} - (m g h_{f} - m g h_{i}) \frac{1}{2} m v_{f}^{2} - \frac{1}{2} m v_{i}^{2} = W_{f} - m g h_{f} + m g h_{i}$ (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 $(v_{i} = 110 km / h)$ , $W_{f}$ is the work done by nonconservative force or the work done by the friction, g is the acceleration due to gravity $(9.8 m / s^{2})$ , $h_{f}$ 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),

$[\frac{1}{2} \times (750 kg) \times {(0)}^{2} - \frac{1}{2} \times (750 kg) \times {(110 km / h)}^{2}] = [0 - (750 kg) \times (9.8 m / s^{2}) \times h_{f} + (750 kg) \times (9.8 m / s^{2}) \times (0)] - \frac{1}{2} \times (750 kg) \times {(30.56 m / s)}^{2} = - (750 kg) \times (9.8 m / s^{2}) \times h_{f} h_{f} = \frac{- \frac{1}{2} \times (750 kg) \times {(30.56 m / s)}^{2}}{- (750 kg) \times (9.8 m / s^{2})} h_{f} = 47.63 m$

Therefore, the car can go up to the height of $47.63 m$ .

The heat energy generated

(b)

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

Putting all known values,

$[\frac{1}{2} \times (750 kg) \times {(0)}^{2} - \frac{1}{2} \times (750 kg) \times {(110 km / h)}^{2}] = [W_{f} - (750 kg) \times (9.8 m / s^{2}) \times (22.0 m) + (750 kg) \times (9.8 m / s^{2}) \times (0)] - \frac{1}{2} \times (750 kg) \times {(30.56 m / s)}^{2} = W_{f} - (750 kg) \times (9.8 m / s^{2}) \times (22.0 m) W_{f} = [(750 kg) \times (9.8 m / s^{2}) \times (22.0 m) - \frac{1}{2} \times (750 kg) \times {(30.56 m / s)}^{2}] = - 188415.74 J$

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

Therefore, the thermal energy produced due to heat is $188415.74 J$ .

The average frictional force

The work done by the frictional force is,

$W_{f} = - F d$ (1.2)

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

The distance travelled is given as,

$d = \frac{h}{\sin (2.5 °)}$

Putting all known values,

$d = \frac{(22.0 m)}{\sin (2.5 °)} = 504.36 m$

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

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

Putting all known values,

$F = - \frac{(- 188415.74 J)}{(504.36 m)} = 373.57 N$

Therefore, the required average frictional force is $373.57 N$

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College Physics (Urone)

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Short Answer

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Conservation of energy

Maximum height attained by the car

The heat energy generated

The average frictional force

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College Physics (Urone)

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