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You are standing on a bathroom scale in an elevator in a tall building. Your mass is 72 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 bathroom scale reads 1036.8 N at t=4.0 s.

Step by step solution

01

Determine Acceleration

To find the reading of the bathroom scale, we need to calculate the net force acting on you. The elevator's velocity function is \[ v(t) = (3.0 \text{ m/s}^2) t + (0.20 \text{ m/s}^3) t^2. \] We first differentiate this function with respect to time to find acceleration: \[ a(t) = \frac{dv(t)}{dt} = 3.0 \text{ m/s}^2 + 2(0.20 \text{ m/s}^3)t. \] At \( t = 4.0 \text{s} \), substitute \( t \) to find the acceleration: \[ a(4.0) = 3.0 + 2(0.20)(4.0) = 3.0 + 1.6 = 4.6 \text{ m/s}^2. \]
02

Calculate Net Force

The net force is the sum of gravitational force and the force due to acceleration. The gravitational force \( F_g \) is computed as \[ F_g = mg = 72 \text{ kg} \times 9.8 \text{ m/s}^2 = 705.6 \text{ N}. \] The additional force due to the elevator's acceleration \( F_a \) is \[ F_a = ma = 72 \text{ kg} \times 4.6 \text{ m/s}^2 = 331.2 \text{ N}. \]
03

Find Scale Reading

The scale reading corresponds to the apparent weight, which is the sum of gravitational and additional forces. Therefore, \[ \text{Scale Reading} = F_g + F_a = 705.6 \text{ N} + 331.2 \text{ N} = 1036.8 \text{ N}. \] This reading reflects both the gravitational force and the force due to the elevator's upward acceleration.

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Key Concepts

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

Acceleration in Elevator Motion
In an elevator scenario, where velocity changes over time, acceleration is a key factor. It tells us how quickly the speed of the elevator is changing. To find acceleration, we use the velocity function, which in this case is given as \(v(t) = (3.0 \, \text{m/s}^2) t + (0.20 \, \text{m/s}^3) t^2\). By differentiating this velocity function with respect to time, we get acceleration.
Doing this gives us \(a(t) = 3.0 \, \text{m/s}^2 + 2(0.20 \, \text{m/s}^3)t\). At a specific time, such as \( t = 4.0 \, \text{s}\), we substitute \(t\) into this equation to find \(a(4.0) = 4.6 \, \text{m/s}^2\).
  • Acceleration is the rate of change of velocity.
  • It helps determine how forces will affect apparent weight in the elevator.
Force Calculation in Dynamics
The net force acting on you in an elevator, or in any physics problem, comprises different components of force. Here, it includes forces due to gravity and the acceleration of the elevator. Calculating these forces requires using Newton's Second Law, \( F = ma\), where \(F\) is the force, \(m\) is mass, and \(a\) is acceleration.
The gravitational force, \(F_g\), is straightforward, calculated using \(g\), the acceleration due to gravity (\(9.8 \, \text{m/s}^2\)). Thus, \(F_g = 72 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 705.6 \, \text{N}\).The additional force due to elevator's acceleration, \(F_a\), employs the total acceleration value: \(F_a = 72 \, \text{kg} \times 4.6 \, \text{m/s}^2 = 331.2 \, \text{N}\).
  • Forces in dynamics are the product of mass and acceleration.
  • Net force combines all relevant forces to determine apparent weight.
Understanding Apparent Weight
Apparent weight is what you "feel" as your weight when in a different accelerating context—like an elevator. This differs from your true weight because of the acceleration affecting the system.
In this problem, your apparent weight is determined by the scale reading, which accounts for both gravitational force and the extra force due to acceleration. Hence, the scale shows \(F_g + F_a\), totaling \(1036.8 \, \text{N}\).
  • Apparent weight changes with acceleration.
  • In an upward accelerating elevator, you feel heavier.
Elevator Physics in Practice
Elevator physics involves understanding how acceleration affects the forces we experience. It exemplifies Newton's laws in a real-world setting.
As the elevator moves, its acceleration alters the tension in the cable and the pressure on the scale. This is a prime example of physics concepts in action, illustrating how net force and resultant acceleration affect motion and perceived weight.
  • Elevators demonstrate Newton's Second Law: \(F = ma\).
  • Real-world applications help visualize theoretical physics concepts.

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Most popular questions from this chapter

On the ride "Spindletop" at the amusement park Six Flags Over Texas, people stood against the inner wall of a hollow vertical cylinder with radius 2.5 \(\mathrm{m}\) . The cylinder started to rotate, and when it reached a constant rotation rate of 0.60 rev \(/ \mathrm{s}\) , the floor on which people were standing dropped about 0.5 \(\mathrm{m}\) . The people remained pinned against the wall. (a) Draw a force diagram for a person on this ride, after the floor has dropped. (b) What minimum coefficient of station is required if the person on the ride is not to slide downward to the new position of the floor? (c) Does your answer in part (b) depend on the mass of the passenger? (Note: When the ride is over, the cylinder is slowly brought to rest. As it slows down, people side down the walls to the floor.)

Losing Cargo. A \(12.0-\mathrm{kg}\) box rests on the flat floor of a truck. The coefficients of friction between the box and floor are \(\mu_{s}=0.19\) and \(\mu_{k}=0.15 .\) The truck stops at a stop sign and then starts to move with an acceleration of 2.20 \(\mathrm{m} / \mathrm{s}^{2} .\) If the box is 1.80 \(\mathrm{m}\) from the rear of the truck when the truck starts, how much time elapses before the box falls off the truck? How far does the truck travel in this time?

Stopping Distance. (a) If the coefficient of kinctic friction between tires and dry pavement is 0.80 , what is the shortest distance in which you can stop an automobile by locking the brakes when traveling at 28.7 \(\mathrm{m} / \mathrm{s}\) (about 65 \(\mathrm{mi} / \mathrm{h} ) ?\) (b) On wet pavement the coefficient of kinetic friction may be only \(0.25 .\) How fast should you drive on wet pavement in order to be able to stop in the same distance as in part (a)? (Note: Locking the brakes is not the safest way to stop.)

A physics student playing with an air hockey table (a frictionless surface) finds that if she gives the puck a velocity of 3.80 \(\mathrm{m} / \mathrm{s}\) along the length \((1.75 \mathrm{m})\) of the table at one end, by the time it has reached the other end the puck has drifted 2.50 \(\mathrm{cm}\) to the right but still has a velocity component along the length of 3.80 \(\mathrm{m} / \mathrm{s} .\) She correctly concludes that the table is not level and correctly calculates its inclination from the given information. What is the angle of inclination?

Apparent Weight. A 550 -N physics student stands on a bathroom scale in an \(850-k g\) (including the student) elevator that is supported by a cable. As the elevator starts moving, the scale reads 450 \(\mathrm{N}\) . (a) Find the acceleration of the elevator (magnitude and direction). (b) What is the acceleration if the scale reads 670 \(\mathrm{N}\) ? (c) If the scale reads zero, should the student worry? Explain. (d) What is the tension in the cable in parts (a) and (c)?

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