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A 75.0 -kg man is riding an escalator in a shopping mall. The escalator moves the man at a constant velocity from ground level to the floor above, a vertical height of \(4.60 \mathrm{m}\). What is the work done on the man by (a) the gravitational force and (b) the escalator?

Short Answer

Expert verified
(a) The gravitational force does -3381 J of work. (b) The escalator does 3381 J of work.

Step by step solution

01

Understanding the Work Done by Gravitational Force

The work done by a gravitational force is calculated by the formula: \( W = F imes d \times \cos(\theta) \). Here, \( F \) is the force (which is the man's weight), \( d \) is the displacement, and \( \theta \) is the angle between the force and displacement, which is 180 degrees when going against gravity.
02

Calculate the Force of Gravity

The force of gravity acting on the man is given by the equation: \( F = m \times g \), where \( m = 75.0 \, \text{kg} \) is the mass of the man and \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity. \( F = 75.0 \times 9.8 = 735 \, \text{N} \).
03

Calculate Work Done by Gravitational Force

Substitute the values into the work formula: \( W_{gravity} = 735 \times 4.60 \times \cos(180^{\circ}) \). Since \( \cos(180^{\circ}) = -1 \), the calculation becomes: \( W_{gravity} = 735 \times 4.60 \times (-1) = -3381 \text{ J} \).
04

Understanding the Work Done by the Escalator

Since the escalator moves the man at a constant velocity vertically upwards, the work done by the escalator is equal in magnitude and opposite in sign to the work done by the gravitational force. This means that the escalator does positive work.
05

Calculate Work Done by the Escalator

Since the magnitude of the work done by the escalator equals the work done against gravity, but with positive sign, we have: \( W_{escalator} = 3381 \text{ J} \).

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Key Concepts

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

Gravitational Force
Gravitational force is a fundamental concept in physics, referring to the attractive force between two masses. On Earth, this force is primarily observed as the weight of an object. The gravitational force acting on an object is calculated using the formula:
  • \( F = m \times g \)
Where \( F \) is the force in newtons (N), \( m \) is the mass in kilograms (kg), and \( g \) is the acceleration due to gravity, approximately \( 9.8 \, \text{m/s}^2 \).
In the context of a 75 kg man, the gravitational force pulling him down is \( 735 \, \text{N} \). This is because the weight—effectively the gravitational force—is the product of his mass and the gravitational acceleration. Understanding this formula helps us calculate the work done by gravity when considering movement in a vertical direction.
Escalator Physics
Escalator physics involves understanding how an escalator facilitates movement, particularly against gravitational pull. An escalator moving a person upward performs mechanical work. This work is achieved through electrical power that drives the escalator's motors.
  • When a person stands on an escalator, they experience a lift from one level to another without any additional effort.
  • The escalator does work equivalent to the gravitational work in magnitude but in opposite direction.
In our example, the escalator enables the man to move a vertical height of 4.60 meters at constant speed, which counteracts gravity. Importantly, this movement is simplified by the concept of constant velocity where speed remains steady, canceling any acceleration, and easing the computational requirements for understanding work done.
Constant Velocity
Constant velocity in physics means that an object's speed and direction remain unchanged over time. In simple terms, if an object moves at a constant velocity, it covers equal distances in equal time intervals, and there is no net acceleration.
  • This is crucial in dynamics as no net force is required to maintain constant velocity.
  • For the escalator, moving a person upwards at constant velocity suggests steady energy input and output balance.
The principle of constant velocity simplifies calculations for work and energy as it implies that the net force acting on the object is zero. Thus, the only forces that need to be considered here are gravity and the force exerted by the escalator opposing it.
Work-Energy Principle
The work-energy principle is a fundamental idea in physics stating that the work done on an object is equal to the change in its kinetic energy. This principle is versatile and applies to scenarios with varied forces and paths.
  • In our context, it helps us understand how energy is transferred while riding an escalator.
  • The work done by the gravitational force is \( -3381 \text{ J} \), indicating a transfer out of the man's potential energy.
  • The escalator compensates this by doing \( 3381 \text{ J} \) of positive work, maintaining the man’s kinetic energy constant.
This principle emphasizes how, despite the gravitational force acting downwards, the escalator's work balances it out, keeping the velocity constant. Thus, energy conservation is maintained, showcasing compelling interrelations among forces, motion, and energy.

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Most popular questions from this chapter

A semitrailer is coasting downhill along a mountain highway when its brakes fail. The driver pulls onto a runaway-truck ramp that is inclined at an angle of \(14.0^{\circ}\) above the horizontal. The semitrailer coasts to a stop after traveling \(154 \mathrm{m}\) along the ramp. What was the truck's initial speed? Neglect air resistance and friction.

A person pulls a toboggan for a distance of \(35.0 \mathrm{m}\) along the snow with a rope directed \(25.0^{\circ}\) above the snow. The tension in the rope is \(94.0 \mathrm{N} .\) (a) How much work is done on the toboggan by the tension force? (b) How much work is done if the same tension is directed parallel to the snow?

A 35-kg girl is bouncing on a trampoline. During a certain interval after she leaves the surface of the trampoline, her kinetic energy decreases to 210 J from 440 J. How high does she rise during this interval? Neglect air resistance.

A 55.0-kg skateboarder starts out with a speed of 1.80 \(\mathrm{m} / \mathrm{s}\). He does \(+80.0 \mathrm{J}\) of work on himself by pushing with his feet against the ground. In addition, friction does -265 J of work on him. In both cases, the forces doing the work are nonconservative. The final speed of the skateboarder is \(6.00 \mathrm{m} / \mathrm{s} .\) (a) Calculate the change \(\left(\Delta \mathrm{PE}=\mathrm{PE}_{\mathrm{f}}-\mathrm{PE}_{0}\right)\) in the gravitational potential energy. (b) How much has the vertical height of the skater changed, and is the skater above or below the starting point?

A person pushes a 16.0-kg shopping cart at a constant velocity for a distance of \(22.0 \mathrm{m}\). She pushes in a direction \(29.0^{\circ}\) below the horizontal. A \(48.0-\mathrm{N}\) frictional force opposes the motion of the cart. (a) What is the magnitude of the force that the shopper exerts? Determine the work done by (b) the pushing force, (c) the frictional force, and (d) the gravitational force.

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