Q53P - University Physics with Modern Physics

Question

The table* gives automobile performance data for a few types of cars:

(a) During an acceleration of 0 to $\mathbf{60} \mathbf{mph}$, which car has the largest average net force acting on it? The smallest? (b) During this acceleration, for which car would the average net force on a  $\mathbf{72}.\mathbf{0}-\mathbf{kg}$passenger be the largest? The smallest? (c) When the Ferrari F430 accelerates from  0to 100 mph in$\mathbf{8}.\mathbf{6} \mathbf{s}$, what is the average net force acting on it? How does this net force compare with the average net force during the acceleration from 0 to $\mathbf{60} \mathbf{mph}$? Explain why these average net forces might differ. (d) Discuss why a car has a top speed. What is the net force on the Ferrari F430 when it is traveling at its top speed, $\mathbf{196} \mathbf{mph}$?

 

Step-by-Step Solution

Answer

 

(a) Ferrari F430 has the largest average net force acting on it.

Alpha Romeo 4C has the smallest average net force acting on it.

(b)Ferrari F430 has the largest average net force on a $72.0 \text{kg}$passenger.

Volvo S60 has the smallest average net force on a $72.0 \text{kg}$passenger.

(c)The average net force acting on FerrariF430 while accelerating 0from 0 to 100mph is $7458.66 \text{N}$.

The average net force from to is lesser than the average net force from to. The reason the forces differs is due to the difference in the acceleration.

(d)Due to the aerodynamic nature of the cars, the car has top speed.  

The net force on the Ferrari when it is travelling at its top speed is 0.

 

Step 1: Identification of the given data

The given data can be listed below as:

• The mass of Alpha Romeo 4C is $m_1=895 \text{kg}$
• The time taken by Alpha Romeo 4C to accelerate from 0 to 60mph is $t_1=4.4 \text{s}$
• The mass of Honda Civic 2.0i is $m_2=1320 \text{kg}$
• The time taken by Honda Civic 2.0i to accelerate from 0 to 60mph is $t_2=6.4 \text{s}$
• The mass of Ferrari F430 is $m_3=1435 \text{kg}$
• The time taken by Ferrari F430 to accelerate from to is.$t_3=3.9 \text{s}$
• The mass of Ford Focus RS500 is $m_4=1468 \text{kg}$
• The time taken by Ford Focus RS500 to accelerate from 0 to 60 mph is $t_4=5.4 \text{s}$
• The mass of Volvo S60 is $m_5=1650 \text{kg}$

• The time taken by Volvo S60 to accelerate from 0 to 60mph is $t_5=7.2 \text{s}$

Step 2: Significance of the net force

The net force is described as the sum of the forces that acts on a particular object. The net force mainly replaces the effect of the original force produced by a particle.

Step 3: (a) Determination of the largest average net force

The equation of the average net force of Alpha Romeo 4C is expressed as:

 $F_1=m_1\frac{v_1}{t_1}$

Here, $F_1$is the average net forceof Alpha Romeo 4C,$m_1$ is the mass of Alpha Romeo 4C,$t_1$ is the taken by Alpha Romeo 4C to accelerate from 0 to 60mph and $v_1$is the velocity of the car.

 

Substitute the values in the above equation.

 $\begin{array}{rcl}{F}_1& =& \left(895 \text{kg}\right)\frac{60 \text{mph}\times \frac{\text{0.44} \text{m/s}}{\text{1} \text{mph}}}{4.4 \text{s}}\\ & =& \left(895 \text{kg}\right)\frac{26.4 \text{m/s}}{4.4 \text{s}}\\ & =& \left(895 \text{kg}\right)6 {\text{m/s}}^2\\ & =& 5370 \text{kg}·{\text{m/s}}^2\\ & & \end{array}$

Hence, further as:

$\begin{array}{rcl}{F}_1& =& 5370 \text{kg}·{\text{m/s}}^2\times \frac{1 \text{N}}{1 \text{kg}·{\text{m/s}}^{\text{2}}}\\ & =& 5370 \text{N}\\ & & \end{array}$

The equation of the average net force of Honda Civic 2.0i is expressed as:

 $F_2=m_2\frac{v_2}{t_2}$ 

Here, $F_2$is the average net forceof Honda Civic 2.0i, is the massof Honda Civic 2.0i,$t_2$  is the taken by Honda Civic 2.0i to accelerate from to and $v_2$ is the velocity of the car.

Substitute the values in the above equation.

 $\begin{array}{rcl}{F}_2& =& \left(1320 \text{kg}\right)\frac{60 \text{mph}\times \frac{\text{0.44} \text{m/s}}{\text{1} \text{mph}}}{6.4 \text{s}}\\ & =& \left(1320 \text{kg}\right)\frac{26.4 \text{m/s}}{6.4 \text{s}}\\ & =& \left(1320 \text{kg}\right)4.125 {\text{m/s}}^2\\ & =& 5445 \text{kg}·{\text{m/s}}^2\\ & & \end{array}$

 

Hence, further as:

 $$\begin{gathered} F_2=5445 \text{kg}·{\text{m/s}}^2\times \frac{1 \text{N}}{1 \text{kg}·{\text{m/s}}^{\text{2}}} \\ =5445 \text{N} \end{gathered}$$ 

The equation of the average net force of Ferrari F430 is expressed as:

 $F_3=m_3\frac{v_3}{t_3}$ 

Here,$F_3$ is the average net force of Ferrari F430, $m_3$ is the mass of Ferrari F430,  $t_3$is the taken by Ferrari F430 to accelerate from to and $v_3$ is the velocity of the car.

 

Substitute the values in the above equation.

 

$\begin{array}{rcl}{F}_3& =& \left(1435 \text{kg}\right)\frac{60 \text{mph}\times \frac{\text{0.44} \text{m/s}}{\text{1} \text{mph}}}{3.9 \text{s}}\\ & =& \left(1435 \text{kg}\right)\frac{26.4 \text{m/s}}{3.9 \text{s}}\\ & =& \left(1435 \text{kg}\right)6.76 {\text{m/s}}^2\\ & =& 9713.84 \text{kg}·{\text{m/s}}^2\\ & & \end{array}$

Hence, further as:

 $$\begin{gathered} F_3=9713.84 \text{kg}·{\text{m/s}}^2\times \frac{1 \text{N}}{1 \text{kg}·{\text{m/s}}^{\text{2}}} \\ =9713.84 \text{N} \end{gathered}$$ 

The equation of the average net force of Ford Focus RS500 is expressed as:

 $F_4=m_4\frac{v_4}{t_4}$

Here,$F_4$ is the average net forceof Ford Focus RS500, $m_4$is the massof Ford Focus RS500,$t_4$ is the taken by Ford Focus RS500 to accelerate from 0 to 60mph and $v_4$ is the velocity of the car.

.

 

Substitute the values in the above equation.

 $\begin{array}{rcl}{F}_4& =& \left(1468 \text{kg}\right)\frac{60 \text{mph}\times \frac{\text{0.44} \text{m/s}}{\text{1} \text{mph}}}{5.4 \text{s}}\\ & =& \left(1468 \text{kg}\right)\frac{26.4 \text{m/s}}{5.4 \text{s}}\\ & =& \left(1468 \text{kg}\right)4.88 {\text{m/s}}^2\\ & =& 7176.88 \text{kg}·{\text{m/s}}^2\\ & & \end{array}$

Hence, further as:

 $\begin{array}{rcl}{F}_4& =& 7176.88 \text{kg}·{\text{m/s}}^2\times \frac{1 \text{N}}{1 \text{kg}·{\text{m/s}}^{\text{2}}}\\ & =& 7176.88 \text{N}\\ & & \end{array}$

The equation of the average net force of Volvo S60 is expressed as:

 $F_5=m_5\frac{v_5}{t_5}$ 

Here, $F_5$ is the average net force of Volvo S60, $m_5$is the mass of Volvo S60, $t_5$is the taken by Volvo S60to accelerate from 0 to 60mph  and $v_5$ is the velocity of the car.

 

Substitute the values in the above equation.

 $\begin{array}{rcl}{F}_5& =& \left(1650 \text{kg}\right)\frac{60 \text{mph}\times \frac{\text{0.44} \text{m/s}}{\text{1} \text{mph}}}{7.2 \text{s}}\\ & =& \left(1650 \text{kg}\right)\frac{26.4 \text{m/s}}{7.2 \text{s}}\\ & =& \left(1650 \text{kg}\right)3.6 {\text{m/s}}^2\\ & =& 6050 \text{kg}·{\text{m/s}}^2\end{array}$

Hence, further as:

 $\begin{array}{rcl}{F}_5& =& 6050 \text{kg}·{\text{m/s}}^2\times \frac{1 \text{N}}{1 \text{kg}·{\text{m/s}}^{\text{2}}}\\ & =& 6050 \text{N}\\ & & \end{array}$

 Thus, Ferrari F430 has the largest average net force acting on it.

Step 4: (a) Determination of the smallest average net force

According to the calculation made in the above step, it can be observed that Alpha Romeo 4C has the smallest average net force.

 

Thus,Alpha Romeo 4C has the smallest average net force acting on it.

 

Step 5: (b) Determination of the largest average net force on a passenger

The equation of the average net force of Alpha Romeo 4C is expressed as:

 $F_1=m\frac{v_1}{t_1}$

Here,  $F_1$is the average net forceof Alpha Romeo 4C, $t_1$is the massof the passenger, is the taken by Alpha Romeo 4C to accelerate from 0 to 60mph and$v_1$is the velocity of the car.

 

Substitute the values in the above equation.

 $\begin{array}{rcl}{F}_1& =& \left(72 \text{kg}\right)\frac{60 \text{mph}\times \frac{\text{0.44} \text{m/s}}{\text{1} \text{mph}}}{4.4 \text{s}}\\ & =& \left(72 \text{kg}\right)\frac{26.4 \text{m/s}}{4.4 \text{s}}\\ & =& \left(72 \text{kg}\right)6 {\text{m/s}}^2\\ & =& 432 \text{kg}·{\text{m/s}}^2\\ & & \end{array}$

Hence, further as:

 $\begin{array}{rcl}{F}_1& =& 432 \text{kg}·{\text{m/s}}^2\times \frac{1 \text{N}}{1 \text{kg}·{\text{m/s}}^{\text{2}}}\\ & =& 432 \text{N}\\ & & \end{array}$

The equation of the average net force of Honda Civic 2.0i is expressed as:

 $F_2=m\frac{v_2}{t_2}$

Here, $F_2$is the average net forceof Honda Civic 2.0i, m is the massof the passenger,  $t_2$is the taken by Honda Civic 2.0i to accelerate from 0 to 60mph and $v_2$ is the velocity of the car.

 

Substitute the values in the above equation.

 $\begin{array}{rcl}{F}_2& =& \left(72 \text{kg}\right)\frac{60 \text{mph}\times \frac{\text{0.44} \text{m/s}}{\text{1} \text{mph}}}{6.4 \text{s}}\\ & =& \left(72 \text{kg}\right)\frac{26.4 \text{m/s}}{6.4 \text{s}}\\ & =& \left(72 \text{kg}\right)4.125 {\text{m/s}}^2\\ & =& 297 \text{kg}·{\text{m/s}}^2\\ & & \end{array}$

Hence, further as:

 $$\begin{gathered} F_2=297 \text{kg}·{\text{m/s}}^2\times \frac{1 \text{N}}{1 \text{kg}·{\text{m/s}}^{\text{2}}} \\ =297 \text{N} \end{gathered}$$

The equation of the average net force of Ferrari F430is expressed as:

 $F_3=m\frac{v_3}{t_3}$

Here,$F_3$ is the average net force of Ferrari F430, m is the mass of the passenger, $t_3$is the taken by Ferrari F430 to accelerate from 0  to 60mph  and $v_3$ is the velocity of the car.

 

Substitute the values in the above equation.

$\begin{array}{rcl}{F}_3& =& \left(72 \text{kg}\right)\frac{60 \text{mph}\times \frac{\text{0.44} \text{m/s}}{\text{1} \text{mph}}}{3.9 \text{s}}\\ & =& \left(72 \text{kg}\right)\frac{26.4 \text{m/s}}{3.9 \text{s}}\\ & =& \left(72 \text{kg}\right)6.76 {\text{m/s}}^2\\ & =& 486.72 \text{kg}·{\text{m/s}}^2\\ & & \end{array}$

Hence, further as 

$$\begin{gathered} F_3=486.72 \text{kg}·{\text{m/s}}^2\times \frac{1 \text{N}}{1 \text{kg}·{\text{m/s}}^{\text{2}}} \\ =486.72 \text{N} \end{gathered}$$

The equation of the average net force of Ford Focus RS500 is expressed as:

 

$F_4=m\frac{v_4}{t_4}$

Here,$F_4$ is the average net forceof Ford Focus RS500, $t_4$  is the taken by Ford Focus RS500 to accelerate from 0 to 60mph  and $v_4$ is the velocity of the car.

Substitute the values in the above equation.

 $\begin{array}{rcl}{F}_4& =& \left(72 \text{kg}\right)\frac{60 \text{mph}\times \frac{\text{0.44} \text{m/s}}{\text{1} \text{mph}}}{5.4 \text{s}}\\ & =& \left(72 \text{kg}\right)\frac{26.4 \text{m/s}}{5.4 \text{s}}\\ & =& \left(72 \text{kg}\right)4.88 {\text{m/s}}^2\\ & =& 351.36 \text{kg}·{\text{m/s}}^2\\ & & \end{array}$

Hence, further as:

 $$\begin{gathered} F_4=351.36 \text{kg}·{\text{m/s}}^2\times \frac{1 \text{N}}{1 \text{kg}·{\text{m/s}}^{\text{2}}} \\ =351.36 \text{N} \end{gathered}$$

The equation of the average net force of Volvo S60 is expressed as:

$F_5=m\frac{v_5}{t_5}$ 

Here, $F_5$ is the average net force of Volvo S60, $t_5$ is the taken by Volvo S60to accelerate from 0 to 60 mph and $v_5$ is the velocity of the car.

 

Substitute the values in the above equation.

$\begin{array}{rcl}{F}_5& =& \left(72 \text{kg}\right)\frac{60 \text{mph}\times \frac{\text{0.44} \text{m/s}}{\text{1} \text{mph}}}{7.2 \text{s}}\\ & =& \left(72 \text{kg}\right)\frac{26.4 \text{m/s}}{7.2 \text{s}}\\ & =& \left(72 \text{kg}\right)3.6 {\text{m/s}}^2\\ & =& 259.2 \text{kg}·{\text{m/s}}^2\\ & & \end{array}$

Hence, further as:

 $$\begin{gathered} F_5=259.2 \text{kg}·{\text{m/s}}^2\times \frac{1 \text{N}}{1 \text{kg}·{\text{m/s}}^{\text{2}}} \\ =259.2 \text{N} \end{gathered}$$ 

Thus, Ferrari F430 has the largest average net force on a $72.0 \text{kg}$passenger

Step 6: (b) Determination of the smallest average net force on a passenger

According to the calculations done above, it has been identified that the Volvo S60 has the smallest average net force.

 

Thus, Volvo S60 has the smallest average net force on a $72.0 \text{kg}$passenger

Step 7: (c) Determination of the average net force on Ferrari F430

The equation of the average net force of Ferrari F430is expressed as:

 $F=m_3\frac{v}{t}$

Here, F is the average net forceof Ferrari F430, is the taken by Ferrari F430 to accelerate from to and is the velocity of the car.

 

Substitute the values in the above equation.

 $\begin{array}{rcl}F& =& \left(1435 \text{kg}\right)\frac{100 \text{mph}\times \frac{\text{0.447} \text{m/s}}{\text{1} \text{mph}}}{8.6 \text{s}}\\ & =& \left(1435 \text{kg}\right)\frac{44.7 \text{m/s}}{8.6 \text{s}}\\ & =& \left(1435 \text{kg}\right)5.1 {\text{m/s}}^2\\ & =& 7458.66 \text{kg}·{\text{m/s}}^2\\ & & \end{array}$

Hence, further as:

 $$\begin{gathered} F=7458.66 \text{kg}·{\text{m/s}}^2\times \frac{1 \text{N}}{1 \text{kg}·{\text{m/s}}^{\text{2}}} \\ =7458.66 \text{N} \end{gathered}$$

Thus, the average net force acting on FerrariF430while accelerating from 0 to100mph is $7458.66 \text{N}$

Step 8: (c) Determination of the comparison of the average net force on Ferrari F430

According to the calculations made above, it has been identified that the average net force from 0 to 100mph is lesser than the average net force from 0 to 60 mph

 

Thus, the average net force from  0to 100mph  is lesser than the average net force from 0 to 60mph. The reason the forces differs is due to the difference in the acceleration.

Step 9: (d) Determination of the reason a car has top speed

The air resistance significantly increases with the increase in the speed of a car. As the engine of a car mainly generates maximum force of driving, then the speed of the car increases along with the increase in the air resistance.

 

Thus, due to the aerodynamic nature of the cars, the car has top speed.  

Step 10: (d) Determination of the net force on Ferrari F430

When Ferrari F430 is travelling at the top speed, then the speed of the car becomes constant. Hence the acceleration and the force acting on the car becomes zero. 

 

Thus, the net force on the Ferrari when it is travelling at its top speed is 0.