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Chapter 5: Problem 53

CE Predict/Explain Riding in an elevator moving with a (a) \(\mathrm{D} \mathrm{o}\) constant upward acceleration, you begin a game of darts. you have to aim your darts higher than, lower than, or the same as when you play darts on solid ground? (b) Choose the best explanation from among the following: I. The elevator accelerates upward, giving its passengers a greater "effective" acceleration of gravity. II. You have to aim lower to compensate for the upward acceleration of the elevator. III. Since the elevator moves with a constant acceleration. New ton's laws apply within the elevator the same as on the ground.

Short Answer

Expert verified
Aim the darts higher; explanation I is correct.

Step by step solution

01

Understand the Problem

You are throwing darts while in an elevator that is accelerating upwards with a constant acceleration. You need to determine the direction (higher, lower, or same) to aim the darts compared to when you are on solid ground.
02

Newton's Laws and Effective Gravity

Consider how an upward acceleration affects the motion inside the elevator. Using Newton's laws, an upward acceleration adds to the gravitational acceleration, creating a greater effective gravitational force on the dart.
03

Analyze Effective Acceleration

When the elevator accelerates upwards, the effective acceleration becomes the sum of gravitational acceleration ( g ) and the elevator's acceleration ( a_e ). So, the effective acceleration inside the elevator is a_{ ext{effective}} = g + a_e .
04

Predict the Dart's Trajectory

Since the effective gravitational acceleration inside the elevator is greater than normal gravity ( g + a_e > g ), the dart will experience a stronger downward force. Aim the darts higher to account for the stronger gravitational pull and hit the target as you would on the ground.
05

Explanation Selection

Choose the correct explanation that aligns with the analysis from Step 4. Explanation I is correct because it recognizes the increased effective gravitational acceleration 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.

Effective Gravity
In the context of an elevator accelerating upwards, 'effective gravity' is a term that denotes the increased force of gravity perceived inside the elevator. This situation is captured by Newton's Second Law of Motion, which states that force is equal to mass times acceleration (\( F = ma \)). Here, the elevator affects the perceived gravitational pull on objects within it.
  • The actual gravitational acceleration on Earth is approximately 9.8 m/s虏, commonly represented as \( g \).
  • When the elevator accelerates upwards with acceleration \( a_e \), this adds to the gravitational pull.
  • The new effective gravitational force experienced by the dart (or any object) inside the elevator becomes \( g + a_e \).
This means that the dart feels a stronger pull downwards than it would on solid ground, so it experiences a greater effective gravitational force.
Upward Acceleration
Upward acceleration in this scenario refers to the elevator's movement against the direction of gravity. It impacts how things move within the elevator environment due to Newton's laws.
  • This acceleration causes the dart and any object inside to feel heavier. This is because the effective gravitational pull is increased.
  • The elevator's upward motion means that it counteracts the downward pull of natural gravity, essentially adding to it.
  • This is why you must consider this increased force when aiming darts.
Simply put, the dart must contend with a stronger downward force due to the combination of gravity and the elevator's acceleration.
Dart Trajectory
The path that a dart follows, known as its trajectory, is altered when thrown in an accelerating elevator. The stronger effective gravity influenced by the elevator means the dart will drop more rapidly towards the ground. The key to understanding this change lies in adjusting the aiming point when throwing the dart.
  • Effective gravity pulls the dart down more aggressively, altering its usual path.
  • To hit your target accurately, you must aim higher than you would usually do on solid ground.
  • This compensates for the additional downward force from the combined effects of natural gravity and the elevator's upward acceleration.
Thus, understanding and compensating for the dart's modified trajectory is crucial for successful dart throwing in such a dynamic environment.

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Most popular questions from this chapter

ce Predict/Explain A small car collides with a large truck. (a) Is the acceleration experienced by the car greater than, less than, or equal to the acceleration experienced by the truck? (b) Choose the best explanation from among the following: I. The truck exerts a larger force on the car, giving it the greater acceleration. II. Both vehicles experience the same magnitude of force, therefore the lightweight car experiences the greater acceleration. III. The greater force exerted on the truck gives it the greater acceleration.

\- ce Predict/Explain You drop two balls of equal diameter from the same height at the same time. Ball 1 is made of metal and has a greater mass than ball \(2,\) which is made of wood. The upward force due to air resistance is the same for both balls. (a) Is the drop time of ball 1 greater than, less than, or equal to the drop time of ball \(2 ?\) (b) Choose the best explanation from among the following: I. The acceleration of gravity is the same for all objects, regardless of mass. II. The more massive ball is harder to accelerate. III. Air resistance has less effect on the more massive ball.

An air-track cart of mass \(m_{1}=0.14 \mathrm{kg}\) is moving with a speed \(v_{0}=1.3 \mathrm{m} / \mathrm{s}\) to the right when it collides with a cart of mass \(\mathrm{m}_{2}=0.25 \mathrm{kg}\) that is at rest. Each cart has a wad of putty on its bumper, and hence they stick together as a result of their collision. Suppose the average contact force between the carts is \(F=1.5 \mathrm{N}\) during the collision. (a) What is the acceleration of cart \(1 ?\) Give direction and magnitude. (b) What is the acceleration of cart \(2 ?\) Give direction and magnitude. (c) How long does it take for both carts to have the same speed? (Once the carts have the same speed the collision is over and the contact force vanishes.) (d) What is the final speed of the carts, \(v_{\mathrm{f}} ?\) (e) Show that \(m_{1}{ }^{2} v_{0}\) is equal to \(\left(m_{1}+m_{2}\right) v_{1}\). (We shall investigate the significance of this result in Chapter 9.3

On a planet far, far away, an astronaut picks up a rock. The rock has a mass of \(5.00 \mathrm{kg}\), and on this particular planet its weight is \(40.0 \mathrm{N}\). If the astronaut exerts an upward force of \(46.2 \mathrm{N}\) on the rock, what is its acceleration?

The combination of "crumple zones" and air bags/seatbelts might increase the distance over which a person stops in a collision to as great as \(1.00 \mathrm{m}\). What is the magnitude of the force exerted on a \(65.0-\mathrm{kg}\) driver who decelerates from \(18.0 \mathrm{m} / \mathrm{s}\) to \(0.00 \mathrm{m} / \mathrm{s}\) over a distance of \(1.00 \mathrm{m} ?\) \(\mathbf{A} .162 \mathrm{N}\) B. \(585 \mathrm{N}\) C. \(1.05 \times 10^{4} \mathrm{N}\) D. \(2.11 \times 10^{4} \mathrm{N}\)
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