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You are standing on a bathroom scale in an elevator in a tall building. Your mass is 64 \(\mathrm{kg}\) . The elevator starts from rest and travels upward with a speed that varies with time according to \(v(t)=\left(3.0 \mathrm{m} / \mathrm{s}^{2}\right) t+\left(0.20 \mathrm{m} / \mathrm{s}^{3}\right) t^{2} .\) When \(t=4.0 \mathrm{s},\) what is the reading of the bathroom scale?

Short Answer

Expert verified
The scale reads approximately 922.24 N at 4 seconds.

Step by step solution

01

Calculate the Acceleration

To find how the acceleration changes over time, we first need to differentiate the velocity function with respect to time. The given velocity function is:\[ v(t) = 3.0 \frac{\mathrm{m}}{\mathrm{s}^2}t + 0.20 \frac{\mathrm{m}}{\mathrm{s}^3}t^2 \]Differentiating it gives us the acceleration:\[ a(t) = \frac{dv}{dt} = 3.0 \frac{\mathrm{m}}{\mathrm{s}^2} + 2 \times 0.20 \frac{\mathrm{m}}{\mathrm{s}^3}t = 3.0 \frac{\mathrm{m}}{\mathrm{s}^2} + 0.40 \frac{\mathrm{m}}{\mathrm{s}^2}t \]
02

Substitute the Time into the Acceleration

Now, we substitute \(t = 4.0\) seconds into the acceleration function:\[ a(4) = 3.0 \frac{\mathrm{m}}{\mathrm{s}^2} + 0.40 \times 4 \frac{\mathrm{m}}{\mathrm{s}^2} = 3.0 \frac{\mathrm{m}}{\mathrm{s}^2} + 1.6 \frac{\mathrm{m}}{\mathrm{s}^2} = 4.6 \frac{\mathrm{m}}{\mathrm{s}^2} \]
03

Calculate the Apparent Weight

The apparent weight is the normal force exerted by the scale, which can be calculated using the formula:\[ N = m(g + a) \]where \(m = 64\,\mathrm{kg}\) is your mass, \(g = 9.81\,\mathrm{m/s^2}\) is the acceleration due to gravity, and \(a = 4.6\,\mathrm{m/s^2}\) is the calculated acceleration.Substituting these values in, we get:\[ N = 64\,(9.81 + 4.6) = 64 \times 14.41 = 922.24\,\mathrm{N} \]
04

Interpret the Result

The reading on the bathroom scale is the apparent weight, which we calculated to be approximately 922.24 N. This value represents the force exerted by the scale that counteracts your mass and the acceleration of the elevator.

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Key Concepts

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

Newton's Laws
Newton's Laws are the foundation of classical mechanics, providing the framework to understand how objects behave under the influence of forces.
The first law states that an object remains at rest or in uniform motion unless acted upon by a force. When you're on the scale, even though the elevator moves, your state of motion changes due to forces acting upon you. The second law expresses the relationship between force, mass, and acceleration (\( F = ma \)). This is crucial when calculating apparent weight, as it involves forces and acceleration. Newton's third law, stating that every action has an equal and opposite reaction, underlies the interaction of the scale and the body. These laws provide the basis for understanding how acceleration affects your apparent weight in the elevator.
Apparent Weight
Apparent weight refers to the force exerted by a scale on a body, perceiving it as weight. In a stationary state, apparent weight equals true weight (\( W = mg \)). But when you're in an accelerating elevator, this changes.
The apparent weight doesn't match the actual gravitational force due to the additional acceleration. The change varies with the direction:
  • When accelerating upwards, apparent weight increases.
  • When accelerating downwards, apparent weight decreases.
Knowing the net acceleration in the system informs the apparent weight.
Elevator Physics
Elevator physics involves scenarios where the acceleration directly impacts apparent weight. When you're in an elevator, forces interact, combining gravitational force with additional forces due to elevator movement.
Depending on whether the elevator speeds up or slows down, your weight reading on the scale could falsely indicate more or less force due to acceleration or deceleration.
  • An upward accelerating elevator increases scale reading.
  • A downward accelerating elevator produces a lighter scale reading.
Understanding this aspect of Newtonian physics gives insight into the reality of riding an elevator.
Kinematics
Kinematics, the study of motion without considering forces, is essential in elevator physics. It focuses on describing motion through variables like velocity and acceleration. In our exercise, the velocity function is described as varying with time.
Exploration of kinematics helps us predict future positions and velocities by examining
  • Initial states and their effects over time
  • Instantaneous velocity and acceleration, derived from differentiating motion equations
It allows for precise control and prediction of elevator movement, providing insight into how motion equations relate to real-world scenarios.
Acceleration Calculation
Calculating acceleration is key to understanding how forces interact in dynamics. From our exercise, we use the derivative of the velocity function:\[ a(t) = \frac{dv}{dt} = 3.0 \frac{\mathrm{m}}{\mathrm{s}^2} + 0.40 \frac{\mathrm{m}}{\mathrm{s}^2}t \]This formula shows how acceleration varies at any given time point. By substituting specific time values, exact acceleration at those times is found.
  • Differentiate the velocity function to find acceleration
  • Substitute desired time values into the derived function to get acceleration at specific moments
  • This gives a clear picture of how quickly the elevator is moving and changing speed
Mastering acceleration calculations ensures a complete grasp of motion dynamics in complex systems.

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Most popular questions from this chapter

A 70 -kg person rides in a 30 -kg cart moving at 12 \(\mathrm{m} / \mathrm{s}\) at the top of a hill that is in the shape of an arc of a circle with a radius of 40 \(\mathrm{m} .\) (a) What is the apparent weight of the person as the cart passes over the top of the hill? (b) Determine the maximum speed that the cart may travel at the top of the hill without losing contact with the surface. Does your answer depend on the mass of the cart or the mass of the person? Explain.

Two 25.0 -N weights are suspended at opposite ends of a rope that passes over a light, frictionless pulley. The pulley is attached to a chain that goes to the ceiling. (a) What is the tension in the rope? (b) What is the tension in the chain?

You are riding in a school bus. As the bus rounds a flat curve at constant speed, a lunch box with mass \(0.500 \mathrm{kg},\) suspended from the ceiling of the bus by a string 1.80 \(\mathrm{m}\) long, is found to hang at rest relative to the bus when the string makes an angle of \(30.0^{\circ}\) with the vertical. In this position the lunch box is 50.0 \(\mathrm{m}\) from the center of curvature of the curve. What is the speed \(v\) of the bus?

A rock with mass \(m=3.00\) kg falls from rest in a viscous medium. The rock is acted on by a net constant downward force of 18.0 \(\mathrm{N}\) (a combination of gravity and the buoyant force exerted by the medium) and by a fluid resistance force \(f=k v\) where \(v\) is the speed in \(\mathrm{m} / \mathrm{s}\) and \(k=2.20 \mathrm{N} \cdot \mathrm{s} / \mathrm{m}(\) see Section 5.3\() .\) (a) Find the initial acceleration \(a_{0 .}\) (b) Find the acceleration when the speed is 3.00 \(\mathrm{m} / \mathrm{s} .\) (c) Find the speed when the acceleration equals 0.1\(a_{0}\) (d) Find the terminal speed \(v_{\mathrm{t}}\) . (e) Find the coordinate, speed, and acceleration 2.00 s after the start of the motion. (f) Find the time required to reach a speed of 0.9\(v_{t} .\)

Two bicycle tires are set rolling with the same initial speed of 3.50 \(\mathrm{m} / \mathrm{s}\) on a long, straight road, and the distance each travels before its speed is reduced by half is measured. One tire is inflated to a pressure of 40 psi and goes 18.1 m; the other is at 105 psi and goes 92.9 \(\mathrm{m} .\) What is the coefficient of rolling friction \(\mu_{\mathrm{r}}\) for each? Assume that the net horizontal force is due to rolling friction only.

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