The magnitude of the combined force of friction at 20 m / s and at 30 m / s .

Question

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Answer 1

Question

Chapter 7, Problem 99P

(a)

To determine

The magnitude of the combined force of friction at $20 m/s$ and at $30 m/s$ .

(a)

Expert Solution

Answer to Problem 99P

The magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

Explanation of Solution

Given:

On a $2.87°$ hill, speed of car is $20 m/s$ ,on a $5.74°$ hill, speed of car is $30 m/s$ and total mass of the car is $1000 kg$ .

Formula used:

Magnitude of force of friction at $20 m/s$ is $F20$ and magnitude of force of friction at $30 m/s$ is $F30$ .

The diagram representing the situation is given below.

Physics for Scientists and Engineers, Chapter 7, Problem 99P , additional homework tip 1

Write the equation of resultant along the incline in equilibrium position.

$∑Fx=0$

Here, $Fx$ is summation of all the forces along incline.

Substitute $mgsinθ−Ff$ for $Fx$ in above equation.

$mgsinθ−Ff=0$

Rearrange above equation.

$Ff=mgsinθ$ …... (1)

Here, $mgsinθ$ is component of $Fg$ along the incline and $Ff$ is the frictional force.

Calculation:

Whenspeed of car is $20 m/s$ .

Substitute $2.87°$ for $θ$ , $1000 kg$ for $m$ and $9.81 m/s2$ for $g$ in equation (1).

$F20=(1000 kg)(9.81 m/s2 )sin(2.87°)≈491 N$

Whenspeed of car is $30 m/s$ .

Substitute $5.74°$ for $θ$ , $1000 kg$ for $m$ and $9.81 m/s2$ for $g$ in equation (1).

$F30=(1000 kg)(9.81 m/s2 )sin(5.74°)≈981 N$

Conclusion:

Thus, the magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

(b)

To determine

The power delivered by the engine to drive the car on a level road at $20 m/s$ and at $30 m/s$ .

(b)

Expert Solution

Answer to Problem 99P

The power delivered by the engine to drive the car on a level road at $20 m/s$ and at $30 m/s$ is $9.8 kW$ and $29 kW$ respectively.

Explanation of Solution

Given:

On a $2.87°$ hill, speed of car is $20 m/s$ , on a $5.74°$ hill, speed of car is $30 m/s$ , total mass of the car is $1000 kg$ the magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

Formula used:

Power delivered by the engine to drive the car on a level road at $20 m/s$ is $P20$ and power delivered by the engine to drive the car on a level road at $30 m/s$ is $P30$ .

Power delivered by the engine to drive the car on a level road in order to overcome friction is calculated by the expression given below.

Write the expression for power delivered.

$P=Ffv$ ........(2)

Here, $P$ is the power delivered and $v$ is the speed of the car.

Calculation:

Substitute $491 N$ for $Ff$ and $20 m/s$ for $v$ in equation (2).

$P20=(491 N)(20 m/s)=9820 W=9.8 kW$

Substitute $981 N$ for $Ff$ and $30 m/s$ for $v$ in equation (2).

$P30=(981 N)(30 m/s)=29430 W=29 kW$

Conclusion:

Thus, the power delivered by the engine to drive the car on a level road at $20 m/s$ and at $30 m/s$ is $9.8 kW$ and $29 kW$ respectively.

(c)

To determine

The angle of the steepest incline up which can be maintained by car at steady speed of $20 m/s$ .

(c)

Expert Solution

Answer to Problem 99P

The angle of the steepest incline up which can be maintained by car at steady speed of $20 m/s$ is $8.8°$ .

Explanation of Solution

Given:

On a $2.87°$ hill, speed of car is $20 m/s$ , on a $5.74°$ hill, speed of car is $30 m/s$ , total mass of the car is $1000 kg$ the magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

Formula used:

The diagram representing the situation is given below.

Physics for Scientists and Engineers, Chapter 7, Problem 99P , additional homework tip 2

Write the equation of resultant along the incline in equilibrium position.

$∑Fx=0$

Here, $Fx$ is summation of all the forces along incline.

Substitute $F−mgsinθ−Ff$ for $Fx$ in above equation.

$F−mgsinθ−Ff=0$

Rearrange above equation.

$F=mgsinθ+Ff$

Here, $F$ is the external force.

Substitute $Pv$ for $F$ in above equation.

$Pv=mgsinθ+Ff$

Rearrange above equation for $θ$ .

$θ=sin−1[Pv−Ffmg]$ …... (3)

Calculation:

Substitute $20 m/s$ for $v$ , $40 kW$ for $P$ , $1000 kg$ for $m$ , $491 N$ for $Ff$ and $9.81 m/s2$ for $g$ in equation (3).

$θ=sin−1[ 40 kW 20 m/s−491 N( 1000 kg)( 9.81 m/s 2 )]=sin−1[15099810]=8.8°$

Conclusion:

Thus, the angle of the steepest incline up which can be maintained by car at steady speed of $20 m/s$ is $8.8°$ .

(d)

To determine

The total useful work in kilometers per liter when engine goes at $30 m/s$ .

(d)

Expert Solution

Answer to Problem 99P

The total useful work in kilometers per liter when engine goes at $30 m/s$ is $6.36 km/L$ .

Explanation of Solution

Given:

The magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

Formula used:

Write the expression for work done by engine.

$W=FΔs$

Here, $W$ is the work done, $F$ is the force and $Δs$ is the distance in kilometers per liter.

From the condition of equivalence:

$F20Δs20=F30Δs30$

Here, $Δs20$ is the distance in kilometers per liter when speed is $20 m/s$ and $Δs30$ is the distance in kilometers per liter when speed is $30 m/s$ .

Rearrange above equation.

$Δs30=F20F30Δs20$ ........(4)

Calculation:

Substitute $491 N$ for $F20$ , $981 N$ for $F30$ and $12.7 km/L$ for $s20$ in equation (4).

$Δs30=491 N981 N(12.7 km/L)=6.36 km/L$

Conclusion:

Thus, the total useful work in kilometers per liter when engine goes at $30 m/s$ is $6.36 km/L$ .

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Answer 2

Question

Chapter 7, Problem 99P

(a)

To determine

The magnitude of the combined force of friction at $20 m/s$ and at $30 m/s$ .

(a)

Expert Solution

Answer to Problem 99P

The magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

Explanation of Solution

Given:

On a $2.87°$ hill, speed of car is $20 m/s$ ,on a $5.74°$ hill, speed of car is $30 m/s$ and total mass of the car is $1000 kg$ .

Formula used:

Magnitude of force of friction at $20 m/s$ is $F20$ and magnitude of force of friction at $30 m/s$ is $F30$ .

The diagram representing the situation is given below.

Physics for Scientists and Engineers, Chapter 7, Problem 99P , additional homework tip 1

Write the equation of resultant along the incline in equilibrium position.

$∑Fx=0$

Here, $Fx$ is summation of all the forces along incline.

Substitute $mgsinθ−Ff$ for $Fx$ in above equation.

$mgsinθ−Ff=0$

Rearrange above equation.

$Ff=mgsinθ$ …... (1)

Here, $mgsinθ$ is component of $Fg$ along the incline and $Ff$ is the frictional force.

Calculation:

Whenspeed of car is $20 m/s$ .

Substitute $2.87°$ for $θ$ , $1000 kg$ for $m$ and $9.81 m/s2$ for $g$ in equation (1).

$F20=(1000 kg)(9.81 m/s2 )sin(2.87°)≈491 N$

Whenspeed of car is $30 m/s$ .

Substitute $5.74°$ for $θ$ , $1000 kg$ for $m$ and $9.81 m/s2$ for $g$ in equation (1).

$F30=(1000 kg)(9.81 m/s2 )sin(5.74°)≈981 N$

Conclusion:

Thus, the magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

(b)

To determine

The power delivered by the engine to drive the car on a level road at $20 m/s$ and at $30 m/s$ .

(b)

Expert Solution

Answer to Problem 99P

The power delivered by the engine to drive the car on a level road at $20 m/s$ and at $30 m/s$ is $9.8 kW$ and $29 kW$ respectively.

Explanation of Solution

Given:

On a $2.87°$ hill, speed of car is $20 m/s$ , on a $5.74°$ hill, speed of car is $30 m/s$ , total mass of the car is $1000 kg$ the magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

Formula used:

Power delivered by the engine to drive the car on a level road at $20 m/s$ is $P20$ and power delivered by the engine to drive the car on a level road at $30 m/s$ is $P30$ .

Power delivered by the engine to drive the car on a level road in order to overcome friction is calculated by the expression given below.

Write the expression for power delivered.

$P=Ffv$ ........(2)

Here, $P$ is the power delivered and $v$ is the speed of the car.

Calculation:

Substitute $491 N$ for $Ff$ and $20 m/s$ for $v$ in equation (2).

$P20=(491 N)(20 m/s)=9820 W=9.8 kW$

Substitute $981 N$ for $Ff$ and $30 m/s$ for $v$ in equation (2).

$P30=(981 N)(30 m/s)=29430 W=29 kW$

Conclusion:

Thus, the power delivered by the engine to drive the car on a level road at $20 m/s$ and at $30 m/s$ is $9.8 kW$ and $29 kW$ respectively.

(c)

To determine

The angle of the steepest incline up which can be maintained by car at steady speed of $20 m/s$ .

(c)

Expert Solution

Answer to Problem 99P

The angle of the steepest incline up which can be maintained by car at steady speed of $20 m/s$ is $8.8°$ .

Explanation of Solution

Given:

On a $2.87°$ hill, speed of car is $20 m/s$ , on a $5.74°$ hill, speed of car is $30 m/s$ , total mass of the car is $1000 kg$ the magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

Formula used:

The diagram representing the situation is given below.

Physics for Scientists and Engineers, Chapter 7, Problem 99P , additional homework tip 2

Write the equation of resultant along the incline in equilibrium position.

$∑Fx=0$

Here, $Fx$ is summation of all the forces along incline.

Substitute $F−mgsinθ−Ff$ for $Fx$ in above equation.

$F−mgsinθ−Ff=0$

Rearrange above equation.

$F=mgsinθ+Ff$

Here, $F$ is the external force.

Substitute $Pv$ for $F$ in above equation.

$Pv=mgsinθ+Ff$

Rearrange above equation for $θ$ .

$θ=sin−1[Pv−Ffmg]$ …... (3)

Calculation:

Substitute $20 m/s$ for $v$ , $40 kW$ for $P$ , $1000 kg$ for $m$ , $491 N$ for $Ff$ and $9.81 m/s2$ for $g$ in equation (3).

$θ=sin−1[ 40 kW 20 m/s−491 N( 1000 kg)( 9.81 m/s 2 )]=sin−1[15099810]=8.8°$

Conclusion:

Thus, the angle of the steepest incline up which can be maintained by car at steady speed of $20 m/s$ is $8.8°$ .

(d)

To determine

The total useful work in kilometers per liter when engine goes at $30 m/s$ .

(d)

Expert Solution

Answer to Problem 99P

The total useful work in kilometers per liter when engine goes at $30 m/s$ is $6.36 km/L$ .

Explanation of Solution

Given:

The magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

Formula used:

Write the expression for work done by engine.

$W=FΔs$

Here, $W$ is the work done, $F$ is the force and $Δs$ is the distance in kilometers per liter.

From the condition of equivalence:

$F20Δs20=F30Δs30$

Here, $Δs20$ is the distance in kilometers per liter when speed is $20 m/s$ and $Δs30$ is the distance in kilometers per liter when speed is $30 m/s$ .

Rearrange above equation.

$Δs30=F20F30Δs20$ ........(4)

Calculation:

Substitute $491 N$ for $F20$ , $981 N$ for $F30$ and $12.7 km/L$ for $s20$ in equation (4).

$Δs30=491 N981 N(12.7 km/L)=6.36 km/L$

Conclusion:

Thus, the total useful work in kilometers per liter when engine goes at $30 m/s$ is $6.36 km/L$ .

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Answer 3

Question

Chapter 7, Problem 99P

(a)

To determine

The magnitude of the combined force of friction at $20 m/s$ and at $30 m/s$ .

(a)

Expert Solution

Answer to Problem 99P

The magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

Explanation of Solution

Given:

On a $2.87°$ hill, speed of car is $20 m/s$ ,on a $5.74°$ hill, speed of car is $30 m/s$ and total mass of the car is $1000 kg$ .

Formula used:

Magnitude of force of friction at $20 m/s$ is $F20$ and magnitude of force of friction at $30 m/s$ is $F30$ .

The diagram representing the situation is given below.

Physics for Scientists and Engineers, Chapter 7, Problem 99P , additional homework tip 1

Write the equation of resultant along the incline in equilibrium position.

$∑Fx=0$

Here, $Fx$ is summation of all the forces along incline.

Substitute $mgsinθ−Ff$ for $Fx$ in above equation.

$mgsinθ−Ff=0$

Rearrange above equation.

$Ff=mgsinθ$ …... (1)

Here, $mgsinθ$ is component of $Fg$ along the incline and $Ff$ is the frictional force.

Calculation:

Whenspeed of car is $20 m/s$ .

Substitute $2.87°$ for $θ$ , $1000 kg$ for $m$ and $9.81 m/s2$ for $g$ in equation (1).

$F20=(1000 kg)(9.81 m/s2 )sin(2.87°)≈491 N$

Whenspeed of car is $30 m/s$ .

Substitute $5.74°$ for $θ$ , $1000 kg$ for $m$ and $9.81 m/s2$ for $g$ in equation (1).

$F30=(1000 kg)(9.81 m/s2 )sin(5.74°)≈981 N$

Conclusion:

Thus, the magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

(b)

To determine

The power delivered by the engine to drive the car on a level road at $20 m/s$ and at $30 m/s$ .

(b)

Expert Solution

Answer to Problem 99P

The power delivered by the engine to drive the car on a level road at $20 m/s$ and at $30 m/s$ is $9.8 kW$ and $29 kW$ respectively.

Explanation of Solution

Given:

On a $2.87°$ hill, speed of car is $20 m/s$ , on a $5.74°$ hill, speed of car is $30 m/s$ , total mass of the car is $1000 kg$ the magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

Formula used:

Power delivered by the engine to drive the car on a level road at $20 m/s$ is $P20$ and power delivered by the engine to drive the car on a level road at $30 m/s$ is $P30$ .

Power delivered by the engine to drive the car on a level road in order to overcome friction is calculated by the expression given below.

Write the expression for power delivered.

$P=Ffv$ ........(2)

Here, $P$ is the power delivered and $v$ is the speed of the car.

Calculation:

Substitute $491 N$ for $Ff$ and $20 m/s$ for $v$ in equation (2).

$P20=(491 N)(20 m/s)=9820 W=9.8 kW$

Substitute $981 N$ for $Ff$ and $30 m/s$ for $v$ in equation (2).

$P30=(981 N)(30 m/s)=29430 W=29 kW$

Conclusion:

Thus, the power delivered by the engine to drive the car on a level road at $20 m/s$ and at $30 m/s$ is $9.8 kW$ and $29 kW$ respectively.

(c)

To determine

The angle of the steepest incline up which can be maintained by car at steady speed of $20 m/s$ .

(c)

Expert Solution

Answer to Problem 99P

The angle of the steepest incline up which can be maintained by car at steady speed of $20 m/s$ is $8.8°$ .

Explanation of Solution

Given:

On a $2.87°$ hill, speed of car is $20 m/s$ , on a $5.74°$ hill, speed of car is $30 m/s$ , total mass of the car is $1000 kg$ the magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

Formula used:

The diagram representing the situation is given below.

Physics for Scientists and Engineers, Chapter 7, Problem 99P , additional homework tip 2

Write the equation of resultant along the incline in equilibrium position.

$∑Fx=0$

Here, $Fx$ is summation of all the forces along incline.

Substitute $F−mgsinθ−Ff$ for $Fx$ in above equation.

$F−mgsinθ−Ff=0$

Rearrange above equation.

$F=mgsinθ+Ff$

Here, $F$ is the external force.

Substitute $Pv$ for $F$ in above equation.

$Pv=mgsinθ+Ff$

Rearrange above equation for $θ$ .

$θ=sin−1[Pv−Ffmg]$ …... (3)

Calculation:

Substitute $20 m/s$ for $v$ , $40 kW$ for $P$ , $1000 kg$ for $m$ , $491 N$ for $Ff$ and $9.81 m/s2$ for $g$ in equation (3).

$θ=sin−1[ 40 kW 20 m/s−491 N( 1000 kg)( 9.81 m/s 2 )]=sin−1[15099810]=8.8°$

Conclusion:

Thus, the angle of the steepest incline up which can be maintained by car at steady speed of $20 m/s$ is $8.8°$ .

(d)

To determine

The total useful work in kilometers per liter when engine goes at $30 m/s$ .

(d)

Expert Solution

Answer to Problem 99P

The total useful work in kilometers per liter when engine goes at $30 m/s$ is $6.36 km/L$ .

Explanation of Solution

Given:

The magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

Formula used:

Write the expression for work done by engine.

$W=FΔs$

Here, $W$ is the work done, $F$ is the force and $Δs$ is the distance in kilometers per liter.

From the condition of equivalence:

$F20Δs20=F30Δs30$

Here, $Δs20$ is the distance in kilometers per liter when speed is $20 m/s$ and $Δs30$ is the distance in kilometers per liter when speed is $30 m/s$ .

Rearrange above equation.

$Δs30=F20F30Δs20$ ........(4)

Calculation:

Substitute $491 N$ for $F20$ , $981 N$ for $F30$ and $12.7 km/L$ for $s20$ in equation (4).

$Δs30=491 N981 N(12.7 km/L)=6.36 km/L$

Conclusion:

Thus, the total useful work in kilometers per liter when engine goes at $30 m/s$ is $6.36 km/L$ .

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Answer 4

Question

Chapter 7, Problem 99P

(a)

To determine

The magnitude of the combined force of friction at $20 m/s$ and at $30 m/s$ .

(a)

Expert Solution

Answer to Problem 99P

The magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

Explanation of Solution

Given:

On a $2.87°$ hill, speed of car is $20 m/s$ ,on a $5.74°$ hill, speed of car is $30 m/s$ and total mass of the car is $1000 kg$ .

Formula used:

Magnitude of force of friction at $20 m/s$ is $F20$ and magnitude of force of friction at $30 m/s$ is $F30$ .

The diagram representing the situation is given below.

Physics for Scientists and Engineers, Chapter 7, Problem 99P , additional homework tip 1

Write the equation of resultant along the incline in equilibrium position.

$∑Fx=0$

Here, $Fx$ is summation of all the forces along incline.

Substitute $mgsinθ−Ff$ for $Fx$ in above equation.

$mgsinθ−Ff=0$

Rearrange above equation.

$Ff=mgsinθ$ …... (1)

Here, $mgsinθ$ is component of $Fg$ along the incline and $Ff$ is the frictional force.

Calculation:

Whenspeed of car is $20 m/s$ .

Substitute $2.87°$ for $θ$ , $1000 kg$ for $m$ and $9.81 m/s2$ for $g$ in equation (1).

$F20=(1000 kg)(9.81 m/s2 )sin(2.87°)≈491 N$

Whenspeed of car is $30 m/s$ .

Substitute $5.74°$ for $θ$ , $1000 kg$ for $m$ and $9.81 m/s2$ for $g$ in equation (1).

$F30=(1000 kg)(9.81 m/s2 )sin(5.74°)≈981 N$

Conclusion:

Thus, the magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

(b)

To determine

The power delivered by the engine to drive the car on a level road at $20 m/s$ and at $30 m/s$ .

(b)

Expert Solution

Answer to Problem 99P

The power delivered by the engine to drive the car on a level road at $20 m/s$ and at $30 m/s$ is $9.8 kW$ and $29 kW$ respectively.

Explanation of Solution

Given:

On a $2.87°$ hill, speed of car is $20 m/s$ , on a $5.74°$ hill, speed of car is $30 m/s$ , total mass of the car is $1000 kg$ the magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

Formula used:

Power delivered by the engine to drive the car on a level road at $20 m/s$ is $P20$ and power delivered by the engine to drive the car on a level road at $30 m/s$ is $P30$ .

Power delivered by the engine to drive the car on a level road in order to overcome friction is calculated by the expression given below.

Write the expression for power delivered.

$P=Ffv$ ........(2)

Here, $P$ is the power delivered and $v$ is the speed of the car.

Calculation:

Substitute $491 N$ for $Ff$ and $20 m/s$ for $v$ in equation (2).

$P20=(491 N)(20 m/s)=9820 W=9.8 kW$

Substitute $981 N$ for $Ff$ and $30 m/s$ for $v$ in equation (2).

$P30=(981 N)(30 m/s)=29430 W=29 kW$

Conclusion:

Thus, the power delivered by the engine to drive the car on a level road at $20 m/s$ and at $30 m/s$ is $9.8 kW$ and $29 kW$ respectively.

(c)

To determine

The angle of the steepest incline up which can be maintained by car at steady speed of $20 m/s$ .

(c)

Expert Solution

Answer to Problem 99P

The angle of the steepest incline up which can be maintained by car at steady speed of $20 m/s$ is $8.8°$ .

Explanation of Solution

Given:

On a $2.87°$ hill, speed of car is $20 m/s$ , on a $5.74°$ hill, speed of car is $30 m/s$ , total mass of the car is $1000 kg$ the magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

Formula used:

The diagram representing the situation is given below.

Physics for Scientists and Engineers, Chapter 7, Problem 99P , additional homework tip 2

Write the equation of resultant along the incline in equilibrium position.

$∑Fx=0$

Here, $Fx$ is summation of all the forces along incline.

Substitute $F−mgsinθ−Ff$ for $Fx$ in above equation.

$F−mgsinθ−Ff=0$

Rearrange above equation.

$F=mgsinθ+Ff$

Here, $F$ is the external force.

Substitute $Pv$ for $F$ in above equation.

$Pv=mgsinθ+Ff$

Rearrange above equation for $θ$ .

$θ=sin−1[Pv−Ffmg]$ …... (3)

Calculation:

Substitute $20 m/s$ for $v$ , $40 kW$ for $P$ , $1000 kg$ for $m$ , $491 N$ for $Ff$ and $9.81 m/s2$ for $g$ in equation (3).

$θ=sin−1[ 40 kW 20 m/s−491 N( 1000 kg)( 9.81 m/s 2 )]=sin−1[15099810]=8.8°$

Conclusion:

Thus, the angle of the steepest incline up which can be maintained by car at steady speed of $20 m/s$ is $8.8°$ .

(d)

To determine

The total useful work in kilometers per liter when engine goes at $30 m/s$ .

(d)

Expert Solution

Answer to Problem 99P

The total useful work in kilometers per liter when engine goes at $30 m/s$ is $6.36 km/L$ .

Explanation of Solution

Given:

The magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

Formula used:

Write the expression for work done by engine.

$W=FΔs$

Here, $W$ is the work done, $F$ is the force and $Δs$ is the distance in kilometers per liter.

From the condition of equivalence:

$F20Δs20=F30Δs30$

Here, $Δs20$ is the distance in kilometers per liter when speed is $20 m/s$ and $Δs30$ is the distance in kilometers per liter when speed is $30 m/s$ .

Rearrange above equation.

$Δs30=F20F30Δs20$ ........(4)

Calculation:

Substitute $491 N$ for $F20$ , $981 N$ for $F30$ and $12.7 km/L$ for $s20$ in equation (4).

$Δs30=491 N981 N(12.7 km/L)=6.36 km/L$

Conclusion:

Thus, the total useful work in kilometers per liter when engine goes at $30 m/s$ is $6.36 km/L$ .

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Answer 5

Chapter 7, Problem 99P

(a)

To determine

The magnitude of the combined force of friction at $20 m/s$ and at $30 m/s$ .

(a)

Expert Solution

Answer to Problem 99P

The magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

Explanation of Solution

Given:

On a $2.87°$ hill, speed of car is $20 m/s$ ,on a $5.74°$ hill, speed of car is $30 m/s$ and total mass of the car is $1000 kg$ .

Formula used:

Magnitude of force of friction at $20 m/s$ is $F20$ and magnitude of force of friction at $30 m/s$ is $F30$ .

The diagram representing the situation is given below.

Physics for Scientists and Engineers, Chapter 7, Problem 99P , additional homework tip 1

Write the equation of resultant along the incline in equilibrium position.

$∑Fx=0$

Here, $Fx$ is summation of all the forces along incline.

Substitute $mgsinθ−Ff$ for $Fx$ in above equation.

$mgsinθ−Ff=0$

Rearrange above equation.

$Ff=mgsinθ$ …... (1)

Here, $mgsinθ$ is component of $Fg$ along the incline and $Ff$ is the frictional force.

Calculation:

Whenspeed of car is $20 m/s$ .

Substitute $2.87°$ for $θ$ , $1000 kg$ for $m$ and $9.81 m/s2$ for $g$ in equation (1).

$F20=(1000 kg)(9.81 m/s2 )sin(2.87°)≈491 N$

Whenspeed of car is $30 m/s$ .

Substitute $5.74°$ for $θ$ , $1000 kg$ for $m$ and $9.81 m/s2$ for $g$ in equation (1).

$F30=(1000 kg)(9.81 m/s2 )sin(5.74°)≈981 N$

Conclusion:

Thus, the magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

(b)

To determine

The power delivered by the engine to drive the car on a level road at $20 m/s$ and at $30 m/s$ .

(b)

Expert Solution

Answer to Problem 99P

The power delivered by the engine to drive the car on a level road at $20 m/s$ and at $30 m/s$ is $9.8 kW$ and $29 kW$ respectively.

Explanation of Solution

Given:

On a $2.87°$ hill, speed of car is $20 m/s$ , on a $5.74°$ hill, speed of car is $30 m/s$ , total mass of the car is $1000 kg$ the magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

Formula used:

Power delivered by the engine to drive the car on a level road at $20 m/s$ is $P20$ and power delivered by the engine to drive the car on a level road at $30 m/s$ is $P30$ .

Power delivered by the engine to drive the car on a level road in order to overcome friction is calculated by the expression given below.

Write the expression for power delivered.

$P=Ffv$ ........(2)

Here, $P$ is the power delivered and $v$ is the speed of the car.

Calculation:

Substitute $491 N$ for $Ff$ and $20 m/s$ for $v$ in equation (2).

$P20=(491 N)(20 m/s)=9820 W=9.8 kW$

Substitute $981 N$ for $Ff$ and $30 m/s$ for $v$ in equation (2).

$P30=(981 N)(30 m/s)=29430 W=29 kW$

Conclusion:

Thus, the power delivered by the engine to drive the car on a level road at $20 m/s$ and at $30 m/s$ is $9.8 kW$ and $29 kW$ respectively.

(c)

To determine

The angle of the steepest incline up which can be maintained by car at steady speed of $20 m/s$ .

(c)

Expert Solution

Answer to Problem 99P

The angle of the steepest incline up which can be maintained by car at steady speed of $20 m/s$ is $8.8°$ .

Explanation of Solution

Given:

On a $2.87°$ hill, speed of car is $20 m/s$ , on a $5.74°$ hill, speed of car is $30 m/s$ , total mass of the car is $1000 kg$ the magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

Formula used:

The diagram representing the situation is given below.

Physics for Scientists and Engineers, Chapter 7, Problem 99P , additional homework tip 2

Write the equation of resultant along the incline in equilibrium position.

$∑Fx=0$

Here, $Fx$ is summation of all the forces along incline.

Substitute $F−mgsinθ−Ff$ for $Fx$ in above equation.

$F−mgsinθ−Ff=0$

Rearrange above equation.

$F=mgsinθ+Ff$

Here, $F$ is the external force.

Substitute $Pv$ for $F$ in above equation.

$Pv=mgsinθ+Ff$

Rearrange above equation for $θ$ .

$θ=sin−1[Pv−Ffmg]$ …... (3)

Calculation:

Substitute $20 m/s$ for $v$ , $40 kW$ for $P$ , $1000 kg$ for $m$ , $491 N$ for $Ff$ and $9.81 m/s2$ for $g$ in equation (3).

$θ=sin−1[ 40 kW 20 m/s−491 N( 1000 kg)( 9.81 m/s 2 )]=sin−1[15099810]=8.8°$

Conclusion:

Thus, the angle of the steepest incline up which can be maintained by car at steady speed of $20 m/s$ is $8.8°$ .

(d)

To determine

The total useful work in kilometers per liter when engine goes at $30 m/s$ .

(d)

Expert Solution

Answer to Problem 99P

The total useful work in kilometers per liter when engine goes at $30 m/s$ is $6.36 km/L$ .

Explanation of Solution

Given:

The magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

Formula used:

Write the expression for work done by engine.

$W=FΔs$

Here, $W$ is the work done, $F$ is the force and $Δs$ is the distance in kilometers per liter.

From the condition of equivalence:

$F20Δs20=F30Δs30$

Here, $Δs20$ is the distance in kilometers per liter when speed is $20 m/s$ and $Δs30$ is the distance in kilometers per liter when speed is $30 m/s$ .

Rearrange above equation.

$Δs30=F20F30Δs20$ ........(4)

Calculation:

Substitute $491 N$ for $F20$ , $981 N$ for $F30$ and $12.7 km/L$ for $s20$ in equation (4).

$Δs30=491 N981 N(12.7 km/L)=6.36 km/L$

Conclusion:

Thus, the total useful work in kilometers per liter when engine goes at $30 m/s$ is $6.36 km/L$ .

Answer 6

Chapter 7, Problem 99P

(a)

To determine

The magnitude of the combined force of friction at $20 m/s$ and at $30 m/s$ .

(a)

Expert Solution

Answer to Problem 99P

The magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

Explanation of Solution

Given:

On a $2.87°$ hill, speed of car is $20 m/s$ ,on a $5.74°$ hill, speed of car is $30 m/s$ and total mass of the car is $1000 kg$ .

Formula used:

Magnitude of force of friction at $20 m/s$ is $F20$ and magnitude of force of friction at $30 m/s$ is $F30$ .

The diagram representing the situation is given below.

Physics for Scientists and Engineers, Chapter 7, Problem 99P , additional homework tip 1

Write the equation of resultant along the incline in equilibrium position.

$∑Fx=0$

Here, $Fx$ is summation of all the forces along incline.

Substitute $mgsinθ−Ff$ for $Fx$ in above equation.

$mgsinθ−Ff=0$

Rearrange above equation.

$Ff=mgsinθ$ …... (1)

Here, $mgsinθ$ is component of $Fg$ along the incline and $Ff$ is the frictional force.

Calculation:

Whenspeed of car is $20 m/s$ .

Substitute $2.87°$ for $θ$ , $1000 kg$ for $m$ and $9.81 m/s2$ for $g$ in equation (1).

$F20=(1000 kg)(9.81 m/s2 )sin(2.87°)≈491 N$

Whenspeed of car is $30 m/s$ .

Substitute $5.74°$ for $θ$ , $1000 kg$ for $m$ and $9.81 m/s2$ for $g$ in equation (1).

$F30=(1000 kg)(9.81 m/s2 )sin(5.74°)≈981 N$

Conclusion:

Thus, the magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

Answer 7

Chapter 7, Problem 99P

(a)

To determine

The magnitude of the combined force of friction at $20 m/s$ and at $30 m/s$ .

Answer 8

(a)

Expert Solution

Answer to Problem 99P

The magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

On a $2.87°$ hill, speed of car is $20 m/s$ ,on a $5.74°$ hill, speed of car is $30 m/s$ and total mass of the car is $1000 kg$ .

Formula used:

Magnitude of force of friction at $20 m/s$ is $F20$ and magnitude of force of friction at $30 m/s$ is $F30$ .

The diagram representing the situation is given below.

Physics for Scientists and Engineers, Chapter 7, Problem 99P , additional homework tip 1

Write the equation of resultant along the incline in equilibrium position.

$∑Fx=0$

Here, $Fx$ is summation of all the forces along incline.

Substitute $mgsinθ−Ff$ for $Fx$ in above equation.

$mgsinθ−Ff=0$

Rearrange above equation.

$Ff=mgsinθ$ …... (1)

Here, $mgsinθ$ is component of $Fg$ along the incline and $Ff$ is the frictional force.

Calculation:

Whenspeed of car is $20 m/s$ .

Substitute $2.87°$ for $θ$ , $1000 kg$ for $m$ and $9.81 m/s2$ for $g$ in equation (1).

$F20=(1000 kg)(9.81 m/s2 )sin(2.87°)≈491 N$

Whenspeed of car is $30 m/s$ .

Substitute $5.74°$ for $θ$ , $1000 kg$ for $m$ and $9.81 m/s2$ for $g$ in equation (1).

$F30=(1000 kg)(9.81 m/s2 )sin(5.74°)≈981 N$

Conclusion:

Thus, the magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

Answer 13

(b)

To determine

The power delivered by the engine to drive the car on a level road at $20 m/s$ and at $30 m/s$ .

(b)

Expert Solution

Answer to Problem 99P

The power delivered by the engine to drive the car on a level road at $20 m/s$ and at $30 m/s$ is $9.8 kW$ and $29 kW$ respectively.

Explanation of Solution

Given:

On a $2.87°$ hill, speed of car is $20 m/s$ , on a $5.74°$ hill, speed of car is $30 m/s$ , total mass of the car is $1000 kg$ the magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

Formula used:

Power delivered by the engine to drive the car on a level road at $20 m/s$ is $P20$ and power delivered by the engine to drive the car on a level road at $30 m/s$ is $P30$ .

Power delivered by the engine to drive the car on a level road in order to overcome friction is calculated by the expression given below.

Write the expression for power delivered.

$P=Ffv$ ........(2)

Here, $P$ is the power delivered and $v$ is the speed of the car.

Calculation:

Substitute $491 N$ for $Ff$ and $20 m/s$ for $v$ in equation (2).

$P20=(491 N)(20 m/s)=9820 W=9.8 kW$

Substitute $981 N$ for $Ff$ and $30 m/s$ for $v$ in equation (2).

$P30=(981 N)(30 m/s)=29430 W=29 kW$

Conclusion:

Thus, the power delivered by the engine to drive the car on a level road at $20 m/s$ and at $30 m/s$ is $9.8 kW$ and $29 kW$ respectively.

Answer 14

(b)

To determine

The power delivered by the engine to drive the car on a level road at $20 m/s$ and at $30 m/s$ .

Answer 15

(b)

Expert Solution

Answer to Problem 99P

The power delivered by the engine to drive the car on a level road at $20 m/s$ and at $30 m/s$ is $9.8 kW$ and $29 kW$ respectively.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given:

On a $2.87°$ hill, speed of car is $20 m/s$ , on a $5.74°$ hill, speed of car is $30 m/s$ , total mass of the car is $1000 kg$ the magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

Formula used:

Power delivered by the engine to drive the car on a level road at $20 m/s$ is $P20$ and power delivered by the engine to drive the car on a level road at $30 m/s$ is $P30$ .

Power delivered by the engine to drive the car on a level road in order to overcome friction is calculated by the expression given below.

Write the expression for power delivered.

$P=Ffv$ ........(2)

Here, $P$ is the power delivered and $v$ is the speed of the car.

Calculation:

Substitute $491 N$ for $Ff$ and $20 m/s$ for $v$ in equation (2).

$P20=(491 N)(20 m/s)=9820 W=9.8 kW$

Substitute $981 N$ for $Ff$ and $30 m/s$ for $v$ in equation (2).

$P30=(981 N)(30 m/s)=29430 W=29 kW$

Conclusion:

Thus, the power delivered by the engine to drive the car on a level road at $20 m/s$ and at $30 m/s$ is $9.8 kW$ and $29 kW$ respectively.

Answer 20

(c)

To determine

The angle of the steepest incline up which can be maintained by car at steady speed of $20 m/s$ .

(c)

Expert Solution

Answer to Problem 99P

The angle of the steepest incline up which can be maintained by car at steady speed of $20 m/s$ is $8.8°$ .

Explanation of Solution

Given:

On a $2.87°$ hill, speed of car is $20 m/s$ , on a $5.74°$ hill, speed of car is $30 m/s$ , total mass of the car is $1000 kg$ the magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

Formula used:

The diagram representing the situation is given below.

Physics for Scientists and Engineers, Chapter 7, Problem 99P , additional homework tip 2

Write the equation of resultant along the incline in equilibrium position.

$∑Fx=0$

Here, $Fx$ is summation of all the forces along incline.

Substitute $F−mgsinθ−Ff$ for $Fx$ in above equation.

$F−mgsinθ−Ff=0$

Rearrange above equation.

$F=mgsinθ+Ff$

Here, $F$ is the external force.

Substitute $Pv$ for $F$ in above equation.

$Pv=mgsinθ+Ff$

Rearrange above equation for $θ$ .

$θ=sin−1[Pv−Ffmg]$ …... (3)

Calculation:

Substitute $20 m/s$ for $v$ , $40 kW$ for $P$ , $1000 kg$ for $m$ , $491 N$ for $Ff$ and $9.81 m/s2$ for $g$ in equation (3).

$θ=sin−1[ 40 kW 20 m/s−491 N( 1000 kg)( 9.81 m/s 2 )]=sin−1[15099810]=8.8°$

Conclusion:

Thus, the angle of the steepest incline up which can be maintained by car at steady speed of $20 m/s$ is $8.8°$ .

Answer 21

(c)

To determine

The angle of the steepest incline up which can be maintained by car at steady speed of $20 m/s$ .

Answer 22

(c)

Expert Solution

Answer to Problem 99P

The angle of the steepest incline up which can be maintained by car at steady speed of $20 m/s$ is $8.8°$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given:

On a $2.87°$ hill, speed of car is $20 m/s$ , on a $5.74°$ hill, speed of car is $30 m/s$ , total mass of the car is $1000 kg$ the magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

Formula used:

The diagram representing the situation is given below.

Physics for Scientists and Engineers, Chapter 7, Problem 99P , additional homework tip 2

Write the equation of resultant along the incline in equilibrium position.

$∑Fx=0$

Here, $Fx$ is summation of all the forces along incline.

Substitute $F−mgsinθ−Ff$ for $Fx$ in above equation.

$F−mgsinθ−Ff=0$

Rearrange above equation.

$F=mgsinθ+Ff$

Here, $F$ is the external force.

Substitute $Pv$ for $F$ in above equation.

$Pv=mgsinθ+Ff$

Rearrange above equation for $θ$ .

$θ=sin−1[Pv−Ffmg]$ …... (3)

Calculation:

Substitute $20 m/s$ for $v$ , $40 kW$ for $P$ , $1000 kg$ for $m$ , $491 N$ for $Ff$ and $9.81 m/s2$ for $g$ in equation (3).

$θ=sin−1[ 40 kW 20 m/s−491 N( 1000 kg)( 9.81 m/s 2 )]=sin−1[15099810]=8.8°$

Conclusion:

Thus, the angle of the steepest incline up which can be maintained by car at steady speed of $20 m/s$ is $8.8°$ .

Answer 27

(d)

To determine

The total useful work in kilometers per liter when engine goes at $30 m/s$ .

(d)

Expert Solution

Answer to Problem 99P

The total useful work in kilometers per liter when engine goes at $30 m/s$ is $6.36 km/L$ .

Explanation of Solution

Given:

The magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

Formula used:

Write the expression for work done by engine.

$W=FΔs$

Here, $W$ is the work done, $F$ is the force and $Δs$ is the distance in kilometers per liter.

From the condition of equivalence:

$F20Δs20=F30Δs30$

Here, $Δs20$ is the distance in kilometers per liter when speed is $20 m/s$ and $Δs30$ is the distance in kilometers per liter when speed is $30 m/s$ .

Rearrange above equation.

$Δs30=F20F30Δs20$ ........(4)

Calculation:

Substitute $491 N$ for $F20$ , $981 N$ for $F30$ and $12.7 km/L$ for $s20$ in equation (4).

$Δs30=491 N981 N(12.7 km/L)=6.36 km/L$

Conclusion:

Thus, the total useful work in kilometers per liter when engine goes at $30 m/s$ is $6.36 km/L$ .

Answer 28

(d)

To determine

The total useful work in kilometers per liter when engine goes at $30 m/s$ .

Answer 29

(d)

Expert Solution

Answer to Problem 99P

The total useful work in kilometers per liter when engine goes at $30 m/s$ is $6.36 km/L$ .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Given:

The magnitude of the combined force of friction at $20 m/s$ is $491 N$ and at $30 m/s$ is $981 N$ .

Formula used:

Write the expression for work done by engine.

$W=FΔs$

Here, $W$ is the work done, $F$ is the force and $Δs$ is the distance in kilometers per liter.

From the condition of equivalence:

$F20Δs20=F30Δs30$

Here, $Δs20$ is the distance in kilometers per liter when speed is $20 m/s$ and $Δs30$ is the distance in kilometers per liter when speed is $30 m/s$ .