The average power of each elevator’s motor during the acceleration.

Question

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Answer 1

Question

Chapter 9, Problem 82PQ

(a)

To determine

The average power of each elevator’s motor during the acceleration.

(a)

Expert Solution

Answer to Problem 82PQ

The average power of each elevator’s motor during the acceleration is $3.88×104 W$ .

Explanation of Solution

Write the expression for the distance travelled during acceleration.

$Δy=v¯t$ (I)

Here, $Δy$ is the distance travelled by elevator, $v¯$ is the average velocity and $t$ is the time taken to reach maximum velocity.

Write the expression for average velocity.

$v¯=vi+vf2$

Here, $vi$ is the initial velocity and $vf$ is the final velocity.

Substitute $vi+vf2$ for $v¯$ in equation (I) to get $Δy$ .

$Δy=(vi+vf2)t$ (II)

The forces acting on the elevator car are gravitational force and the force applied by elevator motor.

According to work-energy theorem, net work done by gravitational force and elevator motor is equal to change in kinetic energy.

Write expression for the change in kinetic energy.

$ΔK=Wmotor+Wg$ (III)

Here, $Wmotor$ is the work done by motor and $Wg$ is he work done by gravity and $ΔK$ is the change in kinetic energy.

Write the expression for the work done by gravity.

$Wg=mgΔycosθ$

Here, $m$ is the mass of elevator full of passengers, $g$ is the acceleration due to gravity and $θ$ is the angle between gravitational force and displacement of elevator.

Since gravitational force acts in downward direction and elevator moves in upward direction, angle $θ=180°$ .

Substitute $180°$ for $θ$ in above equation to get $Wg$ .

$Wg=mgΔycos180°=−mgΔy$

Write the expression for change in kinetic energy of elevator.

$ΔK=12mvf2−12mvi2$

Initially elevator is at rest. Substitute $0 m/s$ for $vi$ in above equation to get $ΔK$ .

$ΔK=12mvf2$

Substitute $12mvf2$ for $ΔK$ , $−mgΔy$ for $Wg$ in equation (III) to get $Wmotor$ .

$12mvf2=Wmotor+−mgΔyWmotor=12mvf2+mgΔy$ (IV)

Write the expression for the power of the motor.

$P¯=WmotorΔt$ (V)

Here, $P¯$ is the average power of elevator motor during acceleration and $Δt$ is the time taken to do work.

Conclusion:

It is given that mass of elevator full of passenger is $1155 kg$ , final velocity of elevator is $6.10 m/s$ and time taken to reach this velocity is $5.00 s$ .

Substitute $0 m/s$ for $vi$ , $6.10 m/s$ for $vf$ and $5.00 s$ for $t$ in equation (II) to get $Δy$ .

$Δy=(0 m/s+6.10 m/s2)(5.00 s)=15.25 m$

Substitute $1155 kg$ for $m$ , $6.10 m/s$ for $vf$ , $9.81 m/s2$ for $g$ and $15.25 m$ for $Δy$ in equation (IV) to get $Wmotor$ .

$Wmotor=12(1155 kg)(6.10 m/s)2+(1155 kg)(9.81 m/s2)(15.25 m)=1.94×105 J$

Substitute $1.94×105 J$ for $Wpower$ and $5.00 s$ for $Δt$ in equation (V) to get $P¯$ .

$P¯=1.94×105 J5.00 s=3.88×104 W$

Therefore, the average power of each elevator’s motor during the acceleration is $3.88×104 W$ .

(b)

To determine

The average power of each elevator’s motor during the cruising phase of its motion.

(b)

Expert Solution

Answer to Problem 82PQ

The average power of each elevator’s motor during the cruising phase of its motion is $6.88×104 W$ .

Explanation of Solution

The elevator attained a cruising speed of $6.10 m/s$ . After this velocity net force on elevator is zero. That is applied force balances weight of the elevator.

Write the expression for the average power during cruising motion.

$P=Fv$ (VI)

Here, $P$ is the average power of the motor during cruising motion of elevator , $F$ is the applied force and $v$ is the velocity of elevator.

Write the expression for $F$ .

$F=mg$

Conclusion:

Substitute $1155 kg$ for $m$ and $9.81 m/s2$ for $g$ in above equation to get $F$ .

$F=(1150 kg)(9.81 m/s2)=11281.5 N$

Substitute $11281.5 N$ for $F$ and $6.10 m/s$ for $v$ in equation (VI) to get $P$ .

$P=(11281.5 N)(6.10 m/s)=6.88×104 W$

Therefore, The average power of each elevator’s motor during the cruising phase of its motion is $6.88×104 W$ .

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Answer 2

Question

Chapter 9, Problem 82PQ

(a)

To determine

The average power of each elevator’s motor during the acceleration.

(a)

Expert Solution

Answer to Problem 82PQ

The average power of each elevator’s motor during the acceleration is $3.88×104 W$ .

Explanation of Solution

Write the expression for the distance travelled during acceleration.

$Δy=v¯t$ (I)

Here, $Δy$ is the distance travelled by elevator, $v¯$ is the average velocity and $t$ is the time taken to reach maximum velocity.

Write the expression for average velocity.

$v¯=vi+vf2$

Here, $vi$ is the initial velocity and $vf$ is the final velocity.

Substitute $vi+vf2$ for $v¯$ in equation (I) to get $Δy$ .

$Δy=(vi+vf2)t$ (II)

The forces acting on the elevator car are gravitational force and the force applied by elevator motor.

According to work-energy theorem, net work done by gravitational force and elevator motor is equal to change in kinetic energy.

Write expression for the change in kinetic energy.

$ΔK=Wmotor+Wg$ (III)

Here, $Wmotor$ is the work done by motor and $Wg$ is he work done by gravity and $ΔK$ is the change in kinetic energy.

Write the expression for the work done by gravity.

$Wg=mgΔycosθ$

Here, $m$ is the mass of elevator full of passengers, $g$ is the acceleration due to gravity and $θ$ is the angle between gravitational force and displacement of elevator.

Since gravitational force acts in downward direction and elevator moves in upward direction, angle $θ=180°$ .

Substitute $180°$ for $θ$ in above equation to get $Wg$ .

$Wg=mgΔycos180°=−mgΔy$

Write the expression for change in kinetic energy of elevator.

$ΔK=12mvf2−12mvi2$

Initially elevator is at rest. Substitute $0 m/s$ for $vi$ in above equation to get $ΔK$ .

$ΔK=12mvf2$

Substitute $12mvf2$ for $ΔK$ , $−mgΔy$ for $Wg$ in equation (III) to get $Wmotor$ .

$12mvf2=Wmotor+−mgΔyWmotor=12mvf2+mgΔy$ (IV)

Write the expression for the power of the motor.

$P¯=WmotorΔt$ (V)

Here, $P¯$ is the average power of elevator motor during acceleration and $Δt$ is the time taken to do work.

Conclusion:

It is given that mass of elevator full of passenger is $1155 kg$ , final velocity of elevator is $6.10 m/s$ and time taken to reach this velocity is $5.00 s$ .

Substitute $0 m/s$ for $vi$ , $6.10 m/s$ for $vf$ and $5.00 s$ for $t$ in equation (II) to get $Δy$ .

$Δy=(0 m/s+6.10 m/s2)(5.00 s)=15.25 m$

Substitute $1155 kg$ for $m$ , $6.10 m/s$ for $vf$ , $9.81 m/s2$ for $g$ and $15.25 m$ for $Δy$ in equation (IV) to get $Wmotor$ .

$Wmotor=12(1155 kg)(6.10 m/s)2+(1155 kg)(9.81 m/s2)(15.25 m)=1.94×105 J$

Substitute $1.94×105 J$ for $Wpower$ and $5.00 s$ for $Δt$ in equation (V) to get $P¯$ .

$P¯=1.94×105 J5.00 s=3.88×104 W$

Therefore, the average power of each elevator’s motor during the acceleration is $3.88×104 W$ .

(b)

To determine

The average power of each elevator’s motor during the cruising phase of its motion.

(b)

Expert Solution

Answer to Problem 82PQ

The average power of each elevator’s motor during the cruising phase of its motion is $6.88×104 W$ .

Explanation of Solution

The elevator attained a cruising speed of $6.10 m/s$ . After this velocity net force on elevator is zero. That is applied force balances weight of the elevator.

Write the expression for the average power during cruising motion.

$P=Fv$ (VI)

Here, $P$ is the average power of the motor during cruising motion of elevator , $F$ is the applied force and $v$ is the velocity of elevator.

Write the expression for $F$ .

$F=mg$

Conclusion:

Substitute $1155 kg$ for $m$ and $9.81 m/s2$ for $g$ in above equation to get $F$ .

$F=(1150 kg)(9.81 m/s2)=11281.5 N$

Substitute $11281.5 N$ for $F$ and $6.10 m/s$ for $v$ in equation (VI) to get $P$ .

$P=(11281.5 N)(6.10 m/s)=6.88×104 W$

Therefore, The average power of each elevator’s motor during the cruising phase of its motion is $6.88×104 W$ .

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Answer 3

Question

Chapter 9, Problem 82PQ

(a)

To determine

The average power of each elevator’s motor during the acceleration.

(a)

Expert Solution

Answer to Problem 82PQ

The average power of each elevator’s motor during the acceleration is $3.88×104 W$ .

Explanation of Solution

Write the expression for the distance travelled during acceleration.

$Δy=v¯t$ (I)

Here, $Δy$ is the distance travelled by elevator, $v¯$ is the average velocity and $t$ is the time taken to reach maximum velocity.

Write the expression for average velocity.

$v¯=vi+vf2$

Here, $vi$ is the initial velocity and $vf$ is the final velocity.

Substitute $vi+vf2$ for $v¯$ in equation (I) to get $Δy$ .

$Δy=(vi+vf2)t$ (II)

The forces acting on the elevator car are gravitational force and the force applied by elevator motor.

According to work-energy theorem, net work done by gravitational force and elevator motor is equal to change in kinetic energy.

Write expression for the change in kinetic energy.

$ΔK=Wmotor+Wg$ (III)

Here, $Wmotor$ is the work done by motor and $Wg$ is he work done by gravity and $ΔK$ is the change in kinetic energy.

Write the expression for the work done by gravity.

$Wg=mgΔycosθ$

Here, $m$ is the mass of elevator full of passengers, $g$ is the acceleration due to gravity and $θ$ is the angle between gravitational force and displacement of elevator.

Since gravitational force acts in downward direction and elevator moves in upward direction, angle $θ=180°$ .

Substitute $180°$ for $θ$ in above equation to get $Wg$ .

$Wg=mgΔycos180°=−mgΔy$

Write the expression for change in kinetic energy of elevator.

$ΔK=12mvf2−12mvi2$

Initially elevator is at rest. Substitute $0 m/s$ for $vi$ in above equation to get $ΔK$ .

$ΔK=12mvf2$

Substitute $12mvf2$ for $ΔK$ , $−mgΔy$ for $Wg$ in equation (III) to get $Wmotor$ .

$12mvf2=Wmotor+−mgΔyWmotor=12mvf2+mgΔy$ (IV)

Write the expression for the power of the motor.

$P¯=WmotorΔt$ (V)

Here, $P¯$ is the average power of elevator motor during acceleration and $Δt$ is the time taken to do work.

Conclusion:

It is given that mass of elevator full of passenger is $1155 kg$ , final velocity of elevator is $6.10 m/s$ and time taken to reach this velocity is $5.00 s$ .

Substitute $0 m/s$ for $vi$ , $6.10 m/s$ for $vf$ and $5.00 s$ for $t$ in equation (II) to get $Δy$ .

$Δy=(0 m/s+6.10 m/s2)(5.00 s)=15.25 m$

Substitute $1155 kg$ for $m$ , $6.10 m/s$ for $vf$ , $9.81 m/s2$ for $g$ and $15.25 m$ for $Δy$ in equation (IV) to get $Wmotor$ .

$Wmotor=12(1155 kg)(6.10 m/s)2+(1155 kg)(9.81 m/s2)(15.25 m)=1.94×105 J$

Substitute $1.94×105 J$ for $Wpower$ and $5.00 s$ for $Δt$ in equation (V) to get $P¯$ .

$P¯=1.94×105 J5.00 s=3.88×104 W$

Therefore, the average power of each elevator’s motor during the acceleration is $3.88×104 W$ .

(b)

To determine

The average power of each elevator’s motor during the cruising phase of its motion.

(b)

Expert Solution

Answer to Problem 82PQ

The average power of each elevator’s motor during the cruising phase of its motion is $6.88×104 W$ .

Explanation of Solution

The elevator attained a cruising speed of $6.10 m/s$ . After this velocity net force on elevator is zero. That is applied force balances weight of the elevator.

Write the expression for the average power during cruising motion.

$P=Fv$ (VI)

Here, $P$ is the average power of the motor during cruising motion of elevator , $F$ is the applied force and $v$ is the velocity of elevator.

Write the expression for $F$ .

$F=mg$

Conclusion:

Substitute $1155 kg$ for $m$ and $9.81 m/s2$ for $g$ in above equation to get $F$ .

$F=(1150 kg)(9.81 m/s2)=11281.5 N$

Substitute $11281.5 N$ for $F$ and $6.10 m/s$ for $v$ in equation (VI) to get $P$ .

$P=(11281.5 N)(6.10 m/s)=6.88×104 W$

Therefore, The average power of each elevator’s motor during the cruising phase of its motion is $6.88×104 W$ .

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Answer 4

Question

Chapter 9, Problem 82PQ

(a)

To determine

The average power of each elevator’s motor during the acceleration.

(a)

Expert Solution

Answer to Problem 82PQ

The average power of each elevator’s motor during the acceleration is $3.88×104 W$ .

Explanation of Solution

Write the expression for the distance travelled during acceleration.

$Δy=v¯t$ (I)

Here, $Δy$ is the distance travelled by elevator, $v¯$ is the average velocity and $t$ is the time taken to reach maximum velocity.

Write the expression for average velocity.

$v¯=vi+vf2$

Here, $vi$ is the initial velocity and $vf$ is the final velocity.

Substitute $vi+vf2$ for $v¯$ in equation (I) to get $Δy$ .

$Δy=(vi+vf2)t$ (II)

The forces acting on the elevator car are gravitational force and the force applied by elevator motor.

According to work-energy theorem, net work done by gravitational force and elevator motor is equal to change in kinetic energy.

Write expression for the change in kinetic energy.

$ΔK=Wmotor+Wg$ (III)

Here, $Wmotor$ is the work done by motor and $Wg$ is he work done by gravity and $ΔK$ is the change in kinetic energy.

Write the expression for the work done by gravity.

$Wg=mgΔycosθ$

Here, $m$ is the mass of elevator full of passengers, $g$ is the acceleration due to gravity and $θ$ is the angle between gravitational force and displacement of elevator.

Since gravitational force acts in downward direction and elevator moves in upward direction, angle $θ=180°$ .

Substitute $180°$ for $θ$ in above equation to get $Wg$ .

$Wg=mgΔycos180°=−mgΔy$

Write the expression for change in kinetic energy of elevator.

$ΔK=12mvf2−12mvi2$

Initially elevator is at rest. Substitute $0 m/s$ for $vi$ in above equation to get $ΔK$ .

$ΔK=12mvf2$

Substitute $12mvf2$ for $ΔK$ , $−mgΔy$ for $Wg$ in equation (III) to get $Wmotor$ .

$12mvf2=Wmotor+−mgΔyWmotor=12mvf2+mgΔy$ (IV)

Write the expression for the power of the motor.

$P¯=WmotorΔt$ (V)

Here, $P¯$ is the average power of elevator motor during acceleration and $Δt$ is the time taken to do work.

Conclusion:

It is given that mass of elevator full of passenger is $1155 kg$ , final velocity of elevator is $6.10 m/s$ and time taken to reach this velocity is $5.00 s$ .

Substitute $0 m/s$ for $vi$ , $6.10 m/s$ for $vf$ and $5.00 s$ for $t$ in equation (II) to get $Δy$ .

$Δy=(0 m/s+6.10 m/s2)(5.00 s)=15.25 m$

Substitute $1155 kg$ for $m$ , $6.10 m/s$ for $vf$ , $9.81 m/s2$ for $g$ and $15.25 m$ for $Δy$ in equation (IV) to get $Wmotor$ .

$Wmotor=12(1155 kg)(6.10 m/s)2+(1155 kg)(9.81 m/s2)(15.25 m)=1.94×105 J$

Substitute $1.94×105 J$ for $Wpower$ and $5.00 s$ for $Δt$ in equation (V) to get $P¯$ .

$P¯=1.94×105 J5.00 s=3.88×104 W$

Therefore, the average power of each elevator’s motor during the acceleration is $3.88×104 W$ .

(b)

To determine

The average power of each elevator’s motor during the cruising phase of its motion.

(b)

Expert Solution

Answer to Problem 82PQ

The average power of each elevator’s motor during the cruising phase of its motion is $6.88×104 W$ .

Explanation of Solution

The elevator attained a cruising speed of $6.10 m/s$ . After this velocity net force on elevator is zero. That is applied force balances weight of the elevator.

Write the expression for the average power during cruising motion.

$P=Fv$ (VI)

Here, $P$ is the average power of the motor during cruising motion of elevator , $F$ is the applied force and $v$ is the velocity of elevator.

Write the expression for $F$ .

$F=mg$

Conclusion:

Substitute $1155 kg$ for $m$ and $9.81 m/s2$ for $g$ in above equation to get $F$ .

$F=(1150 kg)(9.81 m/s2)=11281.5 N$

Substitute $11281.5 N$ for $F$ and $6.10 m/s$ for $v$ in equation (VI) to get $P$ .

$P=(11281.5 N)(6.10 m/s)=6.88×104 W$

Therefore, The average power of each elevator’s motor during the cruising phase of its motion is $6.88×104 W$ .

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Answer 5

Chapter 9, Problem 82PQ

(a)

To determine

The average power of each elevator’s motor during the acceleration.

(a)

Expert Solution

Answer to Problem 82PQ

The average power of each elevator’s motor during the acceleration is $3.88×104 W$ .

Explanation of Solution

Write the expression for the distance travelled during acceleration.

$Δy=v¯t$ (I)

Here, $Δy$ is the distance travelled by elevator, $v¯$ is the average velocity and $t$ is the time taken to reach maximum velocity.

Write the expression for average velocity.

$v¯=vi+vf2$

Here, $vi$ is the initial velocity and $vf$ is the final velocity.

Substitute $vi+vf2$ for $v¯$ in equation (I) to get $Δy$ .

$Δy=(vi+vf2)t$ (II)

The forces acting on the elevator car are gravitational force and the force applied by elevator motor.

According to work-energy theorem, net work done by gravitational force and elevator motor is equal to change in kinetic energy.

Write expression for the change in kinetic energy.

$ΔK=Wmotor+Wg$ (III)

Here, $Wmotor$ is the work done by motor and $Wg$ is he work done by gravity and $ΔK$ is the change in kinetic energy.

Write the expression for the work done by gravity.

$Wg=mgΔycosθ$

Here, $m$ is the mass of elevator full of passengers, $g$ is the acceleration due to gravity and $θ$ is the angle between gravitational force and displacement of elevator.

Since gravitational force acts in downward direction and elevator moves in upward direction, angle $θ=180°$ .

Substitute $180°$ for $θ$ in above equation to get $Wg$ .

$Wg=mgΔycos180°=−mgΔy$

Write the expression for change in kinetic energy of elevator.

$ΔK=12mvf2−12mvi2$

Initially elevator is at rest. Substitute $0 m/s$ for $vi$ in above equation to get $ΔK$ .

$ΔK=12mvf2$

Substitute $12mvf2$ for $ΔK$ , $−mgΔy$ for $Wg$ in equation (III) to get $Wmotor$ .

$12mvf2=Wmotor+−mgΔyWmotor=12mvf2+mgΔy$ (IV)

Write the expression for the power of the motor.

$P¯=WmotorΔt$ (V)

Here, $P¯$ is the average power of elevator motor during acceleration and $Δt$ is the time taken to do work.

Conclusion:

It is given that mass of elevator full of passenger is $1155 kg$ , final velocity of elevator is $6.10 m/s$ and time taken to reach this velocity is $5.00 s$ .

Substitute $0 m/s$ for $vi$ , $6.10 m/s$ for $vf$ and $5.00 s$ for $t$ in equation (II) to get $Δy$ .

$Δy=(0 m/s+6.10 m/s2)(5.00 s)=15.25 m$

Substitute $1155 kg$ for $m$ , $6.10 m/s$ for $vf$ , $9.81 m/s2$ for $g$ and $15.25 m$ for $Δy$ in equation (IV) to get $Wmotor$ .

$Wmotor=12(1155 kg)(6.10 m/s)2+(1155 kg)(9.81 m/s2)(15.25 m)=1.94×105 J$

Substitute $1.94×105 J$ for $Wpower$ and $5.00 s$ for $Δt$ in equation (V) to get $P¯$ .

$P¯=1.94×105 J5.00 s=3.88×104 W$

Therefore, the average power of each elevator’s motor during the acceleration is $3.88×104 W$ .

(b)

To determine

The average power of each elevator’s motor during the cruising phase of its motion.

(b)

Expert Solution

Answer to Problem 82PQ

The average power of each elevator’s motor during the cruising phase of its motion is $6.88×104 W$ .

Explanation of Solution

The elevator attained a cruising speed of $6.10 m/s$ . After this velocity net force on elevator is zero. That is applied force balances weight of the elevator.

Write the expression for the average power during cruising motion.

$P=Fv$ (VI)

Here, $P$ is the average power of the motor during cruising motion of elevator , $F$ is the applied force and $v$ is the velocity of elevator.

Write the expression for $F$ .

$F=mg$

Conclusion:

Substitute $1155 kg$ for $m$ and $9.81 m/s2$ for $g$ in above equation to get $F$ .

$F=(1150 kg)(9.81 m/s2)=11281.5 N$

Substitute $11281.5 N$ for $F$ and $6.10 m/s$ for $v$ in equation (VI) to get $P$ .

$P=(11281.5 N)(6.10 m/s)=6.88×104 W$

Therefore, The average power of each elevator’s motor during the cruising phase of its motion is $6.88×104 W$ .

Answer 6

Chapter 9, Problem 82PQ

(a)

To determine

The average power of each elevator’s motor during the acceleration.

(a)

Expert Solution

Answer to Problem 82PQ

The average power of each elevator’s motor during the acceleration is $3.88×104 W$ .

Explanation of Solution

Write the expression for the distance travelled during acceleration.

$Δy=v¯t$ (I)

Here, $Δy$ is the distance travelled by elevator, $v¯$ is the average velocity and $t$ is the time taken to reach maximum velocity.

Write the expression for average velocity.

$v¯=vi+vf2$

Here, $vi$ is the initial velocity and $vf$ is the final velocity.

Substitute $vi+vf2$ for $v¯$ in equation (I) to get $Δy$ .

$Δy=(vi+vf2)t$ (II)

The forces acting on the elevator car are gravitational force and the force applied by elevator motor.

According to work-energy theorem, net work done by gravitational force and elevator motor is equal to change in kinetic energy.

Write expression for the change in kinetic energy.

$ΔK=Wmotor+Wg$ (III)

Here, $Wmotor$ is the work done by motor and $Wg$ is he work done by gravity and $ΔK$ is the change in kinetic energy.

Write the expression for the work done by gravity.

$Wg=mgΔycosθ$

Here, $m$ is the mass of elevator full of passengers, $g$ is the acceleration due to gravity and $θ$ is the angle between gravitational force and displacement of elevator.

Since gravitational force acts in downward direction and elevator moves in upward direction, angle $θ=180°$ .

Substitute $180°$ for $θ$ in above equation to get $Wg$ .

$Wg=mgΔycos180°=−mgΔy$

Write the expression for change in kinetic energy of elevator.

$ΔK=12mvf2−12mvi2$

Initially elevator is at rest. Substitute $0 m/s$ for $vi$ in above equation to get $ΔK$ .

$ΔK=12mvf2$

Substitute $12mvf2$ for $ΔK$ , $−mgΔy$ for $Wg$ in equation (III) to get $Wmotor$ .

$12mvf2=Wmotor+−mgΔyWmotor=12mvf2+mgΔy$ (IV)

Write the expression for the power of the motor.

$P¯=WmotorΔt$ (V)

Here, $P¯$ is the average power of elevator motor during acceleration and $Δt$ is the time taken to do work.

Conclusion:

It is given that mass of elevator full of passenger is $1155 kg$ , final velocity of elevator is $6.10 m/s$ and time taken to reach this velocity is $5.00 s$ .

Substitute $0 m/s$ for $vi$ , $6.10 m/s$ for $vf$ and $5.00 s$ for $t$ in equation (II) to get $Δy$ .

$Δy=(0 m/s+6.10 m/s2)(5.00 s)=15.25 m$

Substitute $1155 kg$ for $m$ , $6.10 m/s$ for $vf$ , $9.81 m/s2$ for $g$ and $15.25 m$ for $Δy$ in equation (IV) to get $Wmotor$ .

$Wmotor=12(1155 kg)(6.10 m/s)2+(1155 kg)(9.81 m/s2)(15.25 m)=1.94×105 J$

Substitute $1.94×105 J$ for $Wpower$ and $5.00 s$ for $Δt$ in equation (V) to get $P¯$ .

$P¯=1.94×105 J5.00 s=3.88×104 W$

Therefore, the average power of each elevator’s motor during the acceleration is $3.88×104 W$ .

Answer 7

Chapter 9, Problem 82PQ

(a)

To determine

The average power of each elevator’s motor during the acceleration.

Answer 8

(a)

Expert Solution

Answer to Problem 82PQ

The average power of each elevator’s motor during the acceleration is $3.88×104 W$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Write the expression for the distance travelled during acceleration.

$Δy=v¯t$ (I)

Here, $Δy$ is the distance travelled by elevator, $v¯$ is the average velocity and $t$ is the time taken to reach maximum velocity.

Write the expression for average velocity.

$v¯=vi+vf2$

Here, $vi$ is the initial velocity and $vf$ is the final velocity.

Substitute $vi+vf2$ for $v¯$ in equation (I) to get $Δy$ .

$Δy=(vi+vf2)t$ (II)

The forces acting on the elevator car are gravitational force and the force applied by elevator motor.

According to work-energy theorem, net work done by gravitational force and elevator motor is equal to change in kinetic energy.

Write expression for the change in kinetic energy.

$ΔK=Wmotor+Wg$ (III)

Here, $Wmotor$ is the work done by motor and $Wg$ is he work done by gravity and $ΔK$ is the change in kinetic energy.

Write the expression for the work done by gravity.

$Wg=mgΔycosθ$

Here, $m$ is the mass of elevator full of passengers, $g$ is the acceleration due to gravity and $θ$ is the angle between gravitational force and displacement of elevator.

Since gravitational force acts in downward direction and elevator moves in upward direction, angle $θ=180°$ .

Substitute $180°$ for $θ$ in above equation to get $Wg$ .

$Wg=mgΔycos180°=−mgΔy$

Write the expression for change in kinetic energy of elevator.

$ΔK=12mvf2−12mvi2$

Initially elevator is at rest. Substitute $0 m/s$ for $vi$ in above equation to get $ΔK$ .

$ΔK=12mvf2$

Substitute $12mvf2$ for $ΔK$ , $−mgΔy$ for $Wg$ in equation (III) to get $Wmotor$ .

$12mvf2=Wmotor+−mgΔyWmotor=12mvf2+mgΔy$ (IV)

Write the expression for the power of the motor.

$P¯=WmotorΔt$ (V)

Here, $P¯$ is the average power of elevator motor during acceleration and $Δt$ is the time taken to do work.

Conclusion:

It is given that mass of elevator full of passenger is $1155 kg$ , final velocity of elevator is $6.10 m/s$ and time taken to reach this velocity is $5.00 s$ .

Substitute $0 m/s$ for $vi$ , $6.10 m/s$ for $vf$ and $5.00 s$ for $t$ in equation (II) to get $Δy$ .

$Δy=(0 m/s+6.10 m/s2)(5.00 s)=15.25 m$

Substitute $1155 kg$ for $m$ , $6.10 m/s$ for $vf$ , $9.81 m/s2$ for $g$ and $15.25 m$ for $Δy$ in equation (IV) to get $Wmotor$ .

$Wmotor=12(1155 kg)(6.10 m/s)2+(1155 kg)(9.81 m/s2)(15.25 m)=1.94×105 J$

Substitute $1.94×105 J$ for $Wpower$ and $5.00 s$ for $Δt$ in equation (V) to get $P¯$ .

$P¯=1.94×105 J5.00 s=3.88×104 W$

Therefore, the average power of each elevator’s motor during the acceleration is $3.88×104 W$ .

Answer 13

(b)

To determine

The average power of each elevator’s motor during the cruising phase of its motion.

(b)

Expert Solution

Answer to Problem 82PQ

The average power of each elevator’s motor during the cruising phase of its motion is $6.88×104 W$ .

Explanation of Solution

The elevator attained a cruising speed of $6.10 m/s$ . After this velocity net force on elevator is zero. That is applied force balances weight of the elevator.

Write the expression for the average power during cruising motion.

$P=Fv$ (VI)

Here, $P$ is the average power of the motor during cruising motion of elevator , $F$ is the applied force and $v$ is the velocity of elevator.

Write the expression for $F$ .

$F=mg$

Conclusion:

Substitute $1155 kg$ for $m$ and $9.81 m/s2$ for $g$ in above equation to get $F$ .

$F=(1150 kg)(9.81 m/s2)=11281.5 N$

Substitute $11281.5 N$ for $F$ and $6.10 m/s$ for $v$ in equation (VI) to get $P$ .

$P=(11281.5 N)(6.10 m/s)=6.88×104 W$

Therefore, The average power of each elevator’s motor during the cruising phase of its motion is $6.88×104 W$ .

Answer 14

(b)

To determine

The average power of each elevator’s motor during the cruising phase of its motion.

Answer 15

(b)

Expert Solution

Answer to Problem 82PQ

The average power of each elevator’s motor during the cruising phase of its motion is $6.88×104 W$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The elevator attained a cruising speed of $6.10 m/s$ . After this velocity net force on elevator is zero. That is applied force balances weight of the elevator.

Write the expression for the average power during cruising motion.

$P=Fv$ (VI)

Here, $P$ is the average power of the motor during cruising motion of elevator , $F$ is the applied force and $v$ is the velocity of elevator.

Write the expression for $F$ .

$F=mg$

Conclusion:

Substitute $1155 kg$ for $m$ and $9.81 m/s2$ for $g$ in above equation to get $F$ .

$F=(1150 kg)(9.81 m/s2)=11281.5 N$

Substitute $11281.5 N$ for $F$ and $6.10 m/s$ for $v$ in equation (VI) to get $P$ .

$P=(11281.5 N)(6.10 m/s)=6.88×104 W$

Therefore, The average power of each elevator’s motor during the cruising phase of its motion is $6.88×104 W$ .

Answer 20

Ch. 9.4 - In the three cases shown in Figure 9.11, a force...Ch. 9.6 - Prob. 9.2CE Ch. 9.6 - Prob. 9.3CE Ch. 9.7 - Prob. 9.4CE Ch. 9.7 - Prob. 9.5CE Ch. 9.9 - Prob. 9.6CE Ch. 9 - Pick an isolated system for the following...Ch. 9 - Prob. 2PQ Ch. 9 - Prob. 3PQ Ch. 9 - Prob. 4PQ

The average power of each elevator’s motor during the acceleration.

Concept explainers

The average power of each elevator’s motor during the acceleration.

Answer to Problem 82PQ

Explanation of Solution

The average power of each elevator’s motor during the cruising phase of its motion.

Answer to Problem 82PQ

Explanation of Solution

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