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Chapter 4: Problem 54

Extraterrestrial physics. You have landed on an unknown planet, Newtonia, and want to know what objects will weigh there. You find that when a certain tool is pushed on a frictionless horizontal surface by a 12.0 \(\mathrm{N}\) force, it moves 16.0 \(\mathrm{m}\) in the first 2.00 \(\mathrm{s}\) starting from rest. You next observe that if you release this tool from rest at 10.0 \(\mathrm{m}\) above the ground, it takes 2.58 s to reach the ground. What does the tool weigh on Newtonia and what would it weigh on Earth?

Short Answer

Expert verified
The tool weighs 4.52 N on Newtonia and 14.72 N on Earth.

Step by step solution

01

Calculate Horizontal Acceleration

Use the formula for distance traveled with constant acceleration: \( s = ut + \frac{1}{2} a t^2 \). Here, initial velocity \( u = 0 \), distance \( s = 16.0 \, \text{m} \) and time \( t = 2.00 \, \text{s} \). Thus, \( 16.0 = \frac{1}{2} a (2.00)^2 \). Solving for \( a \), we get \( a = 8.0 \, \text{m/s}^2 \).
02

Calculate Mass of the Tool

From Newton's second law, \( F = ma \), where \( F = 12.0 \, \text{N} \) and \( a = 8.0 \, \text{m/s}^2 \). So, \( m = \frac{F}{a} = \frac{12.0}{8.0} = 1.5 \, \text{kg} \).
03

Calculate Gravitational Acceleration on Newtonia

Using the free-fall equation \( h = \frac{1}{2} g t^2 \), where the height \( h = 10.0 \, \text{m} \) and time \( t = 2.58 \, \text{s} \), we have \( 10.0 = \frac{1}{2} g (2.58)^2 \). Solving for \( g \), \( g = 3.01 \, \text{m/s}^2 \).
04

Calculate Weight on Newtonia

The weight of an object is given by \( W = mg \). Using \( m = 1.5 \, \text{kg} \) and \( g = 3.01 \, \text{m/s}^2 \), \( W = 1.5 \times 3.01 = 4.52 \, \text{N} \).
05

Calculate Weight on Earth

On Earth, gravitational acceleration \( g = 9.81 \, \text{m/s}^2 \). The weight \( W = mg = 1.5 \times 9.81 = 14.72 \, \text{N} \).

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Key Concepts

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

Gravitational Acceleration
Gravitational acceleration is a measure of how fast objects accelerate when they are dropped. It varies from planet to planet and significantly affects the weight of objects. On Earth, the gravitational acceleration is approximately 9.81 meters per second squared (m/s虏). This means that if you drop an object, its velocity will increase by 9.81 m/s every second it is falling.

On the unknown planet Newtonia, we found the gravitational acceleration using a tool. By dropping it from 10.0 meters and observing that it took 2.58 seconds to reach the ground, we could calculate this acceleration as 3.01 m/s虏. This calculation is essential for understanding how forces like weight act differently in various environments.
Constant Acceleration
Constant acceleration occurs when an object's velocity changes by a uniform amount in each equal time period. It's an important concept when studying motion because it simplifies the equations we use to describe movement.

For example, in a frictionless environment on Newtonia, a tool accelerated at a rate of 8.0 meters per second squared. This was determined by pulling the tool with a known force and measuring its travel distance over time. Understanding constant acceleration helps in predicting an object鈥檚 future position and velocity, making it a powerful tool in physics and engineering.
Weight Calculation
To calculate the weight of an object, you need to know two things: its mass, and the gravitational acceleration acting upon it. The weight formula is simply the product of mass and gravitational acceleration, expressed as \( W = mg \).

For example, a tool with a mass of 1.5 kg on Newtonia experiences a gravitational pull of 3.01 m/s虏, so its weight is \( 1.5 \times 3.01 = 4.52 \) newtons. On Earth, because the gravitational acceleration is stronger at 9.81 m/s虏, the same tool would weigh more, specifically 14.72 newtons.
  • Weight is a force. It depends on the gravity of the planet.
  • Even if mass stays constant, different planets imply different weights.
Free-Fall Equation
The free-fall equation is used to determine the time, velocity, or distance an object experiences when falling under gravitational force alone, typically ignoring air resistance. The basic formula is \( h = \frac{1}{2} g t^2 \), where \( h \) is the height, \( g \) is the gravitational acceleration, and \( t \) is the time taken.

By using this equation, astronauts on Newtonia could find the gravity there by measuring how long it took an object to fall from a known height. Knowing how objects fall in free-fall is foundational in physics and helps in designing everything from safety features on vehicles to predicting the behavior of satellites.

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Most popular questions from this chapter

Planet \(\mathbf{X} !\) When venturing forth on Planet \(X,\) you throw a 5.24 kg rock upward at 13.0 \(\mathrm{m} / \mathrm{s}\) and find that it returns to the same level 1.51 \(\mathrm{s}\) later. What does the rock weigh on Planet \(\mathrm{X}\) ?

A woman is standing in an elevator holding her 2.5 -kg briefcase by its handles. Draw a free-body diagram for the briefcase if the elevator is accelerating downward at 1.50 \(\mathrm{m} / \mathrm{s}^{2}\) and calculate the downward pull of the briefcase on the woman's arm while the elevator is accelerating.

A person throws a 2.5 lb stone into the air with an initial upward speed of 15 \(\mathrm{ft} / \mathrm{s}\) . Make a free-body diagram for this stone (a) after it is free of the person's hand and is traveling upward, (b) at its highest point, (c) when it is traveling downward, and (d) while it is being thrown upward, but is still in contact with the person's hand.

An advertisement claims that a particular automobile can "stop on a dime." What net force would actually be necessary to stop a 850 -kg automobile traveling initially at 45.0 \(\mathrm{km} / \mathrm{h}\) in a distance equal to the diameter of a dime, which is 1.8 \(\mathrm{cm} ?\)

A rifle shoots a 4.20 g bullet out of its barrel. The bullet has a muzzle velocity of 965 \(\mathrm{m} / \mathrm{s}\) just as it leaves the barrel. Assuming a constant horizontal acceleration over a distance of 45.0 \(\mathrm{cm}\) starting from rest, with no friction between the bullet and the barrel, (a) what force does the rifle exert on the bullet while it is in the barrel? (b) Draw a free-body diagram of the bullet (i) while it is in the barrel and (ii) just after it has left the barrel. (c) How many \(g^{\prime}\) s of acceleration does the rifle give this bullet? (d) For how long a time is the bullet in the barrel?
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