Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics
Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics
1st Edition
ISBN: 9781305259836
Author: Debora M. Katz
Publisher: Cengage Learning
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Chapter 9, Problem 82PQ

(a)

To determine

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

(a)

Expert Solution
Check Mark

Answer to Problem 82PQ

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

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=12mvf212mvi2

Initially elevator is at rest. Substitute 0m/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 1155kg, final velocity of elevator is 6.10m/s and time taken to reach this velocity is 5.00s.

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

  Δy=(0m/s+6.10m/s2)(5.00s)=15.25m

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

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

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

  P¯=1.94×105J5.00s=3.88×104W

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

(b)

To determine

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

(b)

Expert Solution
Check Mark

Answer to Problem 82PQ

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

Explanation of Solution

The elevator attained a cruising speed of 6.10m/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 1155kg for m and 9.81m/s2 for g in above equation to get F.

  F=(1150kg)(9.81m/s2)=11281.5N

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

P=(11281.5N)(6.10m/s)=6.88×104W

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

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Chapter 9 Solutions

Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics

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