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A standard bathroom scale is placed on an elevator. A \(30 \mathrm{~kg}\) boy enters the elevator on the first floor and steps on the scale. What will the scale read (in newtons) when the elevator begins to accelerate upward at \(0.5 \mathrm{~m} / \mathrm{s}^{2} ?\)

Short Answer

Expert verified
The scale will read 309 N.

Step by step solution

01

Identify Forces Acting on the Boy

The forces involved are the gravitational force (weight) acting downward and the normal force from the scale acting upward. These forces need to be considered to compute what the scale reads.
02

Calculate the Gravitational Force

The gravitational force can be calculated using the formula\[ F_g = m \times g \]where \( m = 30 \text{ kg} \) and \( g = 9.8 \text{ m/s}^2 \). Substitute these values to get:\[ F_g = 30 \times 9.8 = 294 \text{ N} \].
03

Determine the Net Force on the Boy

When the elevator accelerates upward, the net force is given by:\[ F_{net} = m \times a \]where \( a = 0.5 \text{ m/s}^2 \). Using the mass, we find:\[ F_{net} = 30 \times 0.5 = 15 \text{ N} \].
04

Calculate the Scale Reading

The normal force or the scale reading (\( F_n \)) is the sum of the gravitational force and the net upward force:\[ F_n = F_g + F_{net} = 294 + 15 = 309 \text{ N} \].

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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 that refers to the attractive force between two masses. For an object near the Earth's surface, this force depends on the object's mass and the acceleration due to gravity. This force can be calculated using the formula:
\[ F_g = m \times g \]
Where:
  • \( F_g \) is the gravitational force
  • \( m \) is the mass of the object
  • \( g \) is the acceleration due to gravity, approximately \( 9.8 \text{ m/s}^2 \) on Earth
In our exercise, the mass of the boy is given as \( 30 \text{ kg} \). Using the formula, we can calculate the gravitational force acting on him: \( F_g = 30 \times 9.8 = 294 \text{ N} \). This force acts downward, keeping the boy attracted to Earth. Understanding gravitational force helps in analyzing the object's motion, especially when other forces come into play.
Newton's Second Law
Newton's second law of motion is a key principle that explains how the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The law is represented by the equation:
\[ F_{net} = m \times a \]
Where:
  • \( F_{net} \) is the net force applied to the object
  • \( m \) is the mass of the object
  • \( a \) is the acceleration of the object
In the context of our problem, an elevator accelerates upward at \( 0.5 \text{ m/s}^2 \). For the \( 30 \text{ kg} \) boy standing on the scale, the net force due to this acceleration can be computed as \( F_{net} = 30 \times 0.5 = 15 \text{ N} \). This net force affects how much extra force the scale reads from the boy, as it must account for both the gravitational force and the additional force from acceleration.
Elevator Physics
Elevator physics focuses on how different forces act on objects within an accelerating elevator. When an elevator accelerates upwards, as in our example, the experience alters how an object's weight is perceived.
An accelerating elevator exerts additional upward force on objects inside. This makes them feel 'heavier' due to the combined effect of gravity and the elevator's acceleration. To find the force read by the scale inside, we add the gravitational force and the force due to the elevator's acceleration:
  • Gravitational Force: \( 294 \text{ N} \)
  • Additional Force: \( 15 \text{ N} \)
  • Total Force (Scale Reading): \( 294 + 15 = 309 \text{ N} \)
Understanding these forces helps predict what happens to an object's weight during vertical acceleration, which is a common situation in elevator physics.
Normal Force
The concept of normal force is crucial in understanding interactions between surfaces, especially in cases like our elevator scenario. Normal force is the perpendicular force exerted by a surface to support the weight of an object resting on it.
In the case of an elevator accelerating upward, the normal force becomes greater than the gravitational force acting on the object. This is because the normal force must not only counteract gravity, but also provide the additional upward force required to accelerate the object:
For the boy in the elevator, the normal force \( F_n \) is calculated by adding the gravitational force \( F_g \) and the acceleration force \( F_{net} \):
  • \( F_n = F_g + F_{net} = 294 + 15 = 309 \text{ N} \)
This increased normal force is what the scale measures, indicating the boy's apparent weight in the accelerating elevator. The concept is essential for solving many problems involving steady or changing motion, helping us understand the principles of force and equilibrium.

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Most popular questions from this chapter

A man is dragging a trunk up the loading ramp of a mover's truck. (See Figure \(4.37 .)\) The ramp has a slope angle of \(20.0^{\circ},\) and the man pulls upward with a force \(\vec{F}\) of magnitude \(375 \mathrm{~N}\) whose direction makes an angle of \(30.0^{\circ}\) with the ramp. Find the horizontal and vertical components of the force \(\vec{F}\)

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A spacecraft descends vertically near the surface of Planet X. An upward thrust of \(25.0 \mathrm{kN}\) from its engines slows it down at a rate of \(1.20 \mathrm{~m} / \mathrm{s}^{2},\) but it speeds up at a rate of \(0.80 \mathrm{~m} / \mathrm{s}^{2}\) with an upward thrust of \(10.0 \mathrm{kN}\). (a) In each case, what is the direction of the acceleration of the spacecraft? (b) Draw a free-body diagram for the spacecraft. In each case, speeding up or slowing down, what is the direction of the net force on the spacecraft? (c) Apply Newton's second law to each case, slowing down or speeding up, and use this to find the spacecraft's weight near the surface of Planet X.

At the surface of Jupiter's moon Io, the acceleration due to gravity is \(1.81 \mathrm{~m} / \mathrm{s}^{2} .\) If a piece of ice weighs \(44.0 \mathrm{~N}\) at the surface of the earth, (a) what is its mass on the earth's surface? (b) What are its mass and weight on the surface of Io?

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