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Chapter 7: Problem 10

II A \(60 \mathrm{~kg}\) woman steps onto an up-going escalator, which has an incline of \(32^{\circ}\) with respect to the horizontal and is moving at \(0.5 \mathrm{~m} / \mathrm{s}\). The top of the escalator is \(20 \mathrm{~m}\) above the ground level. Calculate how much work is done by (a) the friction force between the woman's feet and the escalator step, (b) gravity, and (c) the normal force on the woman's feet, as she moves from the bottom to the top of the escalator. What is the total work done on the woman as she moves from the bottom to the top?

Short Answer

Expert verified
The total work done on the woman is \(-11760 \text{ J}\).

Step by step solution

01

Calculate the Vertical Distance

First, calculate the vertical distance the woman travels. Given the height of the escalator is 20 m above the ground level, the vertical distance traveled is directly given as 20 m.
02

Calculate Work Done by Friction

Friction force does no work on the escalator because it is always perpendicular to the direction of motion (up the incline). Thus, the work done by friction is zero.
03

Calculate Work Done by Gravity

The work done by gravity is calculated using the formula:\[ W_g = -mgh \]where:- \( m = 60 \text{ kg} \) is the mass of the woman,- \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity,- \( h = 20 \text{ m} \) is the vertical height.Substitute the values:\[ W_g = -(60 \times 9.8 \times 20) = -11760 \text{ J} \]
04

Calculate Work Done by Normal Force

The work done by the normal force is zero because the normal force is perpendicular to the motion of the escalator.
05

Calculate Total Work Done

Add up all the work done by different forces: friction, gravity, and normal force. Since friction and normal force do no work and gravity does \(-11760 \text{ J}\), the total work done is:\[ W_{\text{total}} = 0 + (-11760) + 0 = -11760 \text{ J} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Escalator Physics Problems
When approaching escalator physics problems, it's essential to understand the forces acting on a person as they move along the escalator. Escalators are inclined planes, which means several forces come into play and affect how work and energy factors are calculated. In our scenario, a woman is moving up an escalator inclined at 32掳 with a constant speed of 0.5 m/s.
Here are some key aspects to consider:
  • Direction of Motion: The escalator's incline means that its direction is not horizontal, which affects how we calculate forces like gravity and normal force.
  • Velocity: Knowing the speed of the person on the escalator helps understand the dynamics involved, although it doesn't directly affect work calculations.
  • Forces: Different forces including friction, gravity, and normal force act on the person along the escalator.
Understanding these aspects makes it easier to calculate the work done by each force and the overall work done as the person travels to the top.
Calculating Work Done by Forces
Work is an essential concept in physics, and calculating it requires understanding the direction of forces relative to motion. The fundamental formula for calculating work is: \[ W = F \cdot d \cdot \cos(\theta) \]
  • Friction Force: In the context of this problem, the friction force acts perpendicular to the motion since it primarily prevents slipping. Thus, work done by friction is zero because the angle theta, here 90掳, leads to \( \cos(90掳) = 0 \).
  • Gravity: Gravity does work on the woman as she moves upward. Since the force of gravity acts downward, the formula becomes \( W_g = -mgh \), where \( h \) is the height moved vertically.
  • Normal Force: Despite acting perpendicular to the inclined plane, the normal force also does no work as it doesn't aid in moving the woman along the escalator. The angle between the normal force and the motion is again 90掳.
The calculation shows that the work done by friction and normal force is zero, whereas gravity does negative work, totaling -11760 J.
Inclined Plane Dynamics
Inclined planes introduce interesting dynamics for forces and motion which can be seen in escalators. When moving along an incline:
  • Gravity Components: Instead of acting directly downwards, gravity is divided into parallel and perpendicular components relative to the incline. The parallel component influences the work done as it acts along the direction of where the escalator goes.
  • Normal Force: Perpendicular to the inclined surface, the normal force supports the object's weight but doesn't influence movement along the incline.
  • Net Force: It's crucial to understand that if the escalator moves at constant speed like here, net force along the incline should be zero, so any applied forces must balance out the gravity component along the incline.
The inclined plane dynamics help explain why certain forces, such as the normal force, do no work and how others, such as the gravitational force, affect the calculation of work done.
Work-Energy Principle
The work-energy principle offers a powerful tool for solving physics problems related to energy transfer. It states that the net work done on an object equals the change in its kinetic energy. Although in this problem, the woman moves at a constant speed, her kinetic energy does not change. Hence, all the work done relates to potential energy change.
  • Potential Energy Change: As the woman ascends the escalator, her gravitational potential energy increases because she's moving to a higher elevation. The work done by gravity reflects this change.
  • Net Work Done: In cases like these where the object鈥檚 kinetic energy doesn't change, the work-energy principle simplifies to consider only potential energy: \( W = \Delta U \). Here \( \Delta U \) is the gravitational potential energy change.
  • Total Work Contribution: With friction and normal forces contributing zero work, gravity is the sole contributor to the work-energy dynamics in the problem.
Understanding the work-energy principle enriches solving problems, by focusing on energy changes rather than individual forces, especially when speed remains constant.

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Most popular questions from this chapter

A factory worker moves a \(30.0 \mathrm{~kg}\) crate a distance of \(4.5 \mathrm{~m}\) along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is 0.25 . (a) What magnitude of force must the worker apply? (b) How much work is done on the crate by the worker's push? (c) How much work is done on the crate by friction? (d) How much work is done by the normal force? By gravity? (e) What is the net work done on the crate?

Your friend (mass \(65.0 \mathrm{~kg}\) ) is standing on the ice in the middle of a frozen pond. There is very little friction between her feet and the ice, so she is unable to walk. Fortunately, a light rope is tied around her waist and you stand on the bank holding the other end. You pull on the rope for \(3.00 \mathrm{~s}\) and accelerate your friend from rest to a speed of \(6.00 \mathrm{~m} / \mathrm{s}\) while you remain at rest. What is the average power supplied by the force you applied?

BIO A typical flying insect applies an average force equal to twice its weight during each downward stroke while hovering. Take the mass of the insect to be \(10 \mathrm{~g},\) and assume the wings move an average downward distance of \(1.0 \mathrm{~cm}\) during each stroke. Assuming 100 downward strokes per second, estimate the average power output of the insect.

A constant horizontal pull of \(8.50 \mathrm{~N}\) drags a box along a horizontal floor through a distance of \(17.4 \mathrm{~m}\). (a) How much work does the pull do on the box? (b) Suppose that the same pull is exerted at an angle above the horizontal. If this pull now does \(65.0 \mathrm{~J}\) of work on the box while pulling it through the same distance, what angle does the force make with the horizontal?

A rope is tied to a box and used to pull the box \(1.5 \mathrm{~m}\) along a h zontal floor. The rope makes an angle of \(30^{\circ}\) with the horizontal has a tension of \(5 \mathrm{~N}\). The opposing friction force between the box the floor is \(1 \mathrm{~N}\). How much work does each of the following forces do on the box: (a) gravity, (b) the tension in the rope, (c) friction, and (d) the normal force? What is the total work done on the box?
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