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Chapter 4: Problem 13

A force acts on a \(1.5-\mathrm{kg},\) mass, giving it an acceleration of \(3.0 \mathrm{~m} / \mathrm{s}^{2} .\) (a) If the same force acts on a 2.5 -kg mass, what acceleration would be produced? (b) What is the magnitude of the force?

Short Answer

Expert verified
(a) The acceleration is 1.8 m/s虏. (b) The force magnitude is 4.5 N.

Step by step solution

01

Understanding the relationship between force, mass, and acceleration

The force acting on an object can be calculated using Newton's second law of motion, which states that force (\( F \)) is the product of mass (\( m \)) and acceleration (\( a \)). This is represented by the equation \( F = m \times a \).
02

Calculate the force using the initial conditions

Given the mass \( m = 1.5 \, \text{kg} \) and acceleration \( a = 3.0 \, \text{m/s}^2 \), we can calculate the force:\[ F = 1.5 \, \text{kg} \times 3.0 \, \text{m/s}^2 \]\[ F = 4.5 \, \text{N} \]
03

Use the calculated force for the new mass to find acceleration

Now, using the force \( F = 4.5 \, \text{N} \) calculated in Step 2, find the acceleration for the second mass, \( m = 2.5 \, \text{kg} \), using the formula:\[ a = \frac{F}{m} \]\[ a = \frac{4.5 \, \text{N}}{2.5 \, \text{kg}} \]\[ a = 1.8 \, \text{m/s}^2 \]
04

Finding the magnitude of the force

The magnitude of the force was already determined in Step 2. It is \( 4.5 \, \text{N} \).

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Key Concepts

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

Force Calculation
Newton's Second Law of Motion offers a straightforward approach to understanding how forces influence motion. At its core, the law reveals that force is the result of multiplying mass and acceleration.
In equation form, it is represented as:
  • \( F = m \times a \)
Given information like mass and acceleration, it's simple to calculate the force exerted. Suppose you have a mass of \(1.5 \, \text{kg}\) accelerating at \(3.0 \, \text{m/s}^2\). Plug these values into the formula:
  • \( F = 1.5 \, \text{kg} \times 3.0 \, \text{m/s}^2 \)
  • \( F = 4.5 \, \text{N} \)
This equation and calculation forms the basis of solving many force-related problems in physics.
Mass and Acceleration Relationship
The interplay between mass and acceleration is key in understanding motion dynamics. One should realize that with a constant force, any increase in mass results in a proportional decrease in acceleration. This inverse relationship can be illustrated when a force acts differently on different masses.
In the given exercise, a force of \(4.5 \, \text{N}\) is initially calculated for a \(1.5 \, \text{kg}\) mass. If this same force is applied to a \(2.5 \, \text{kg}\) mass, find the new acceleration by using:
  • \( a = \frac{F}{m} \)
  • \( a = \frac{4.5 \, \text{N}}{2.5 \, \text{kg}} \)
  • \( a = 1.8 \, \text{m/s}^2 \)
This shows how an increased mass results in a reduced acceleration when the same force is applied.
Physics Problem-Solving
When addressing physics problems, breaking them into manageable parts can simplify complex scenarios. Here鈥檚 a structured approach to attack such problems effectively:
  • **Identify Known Variables:** Gather given data such as mass, acceleration, or force.
  • **Apply Newton's Second Law:** Use \( F = m \times a \) to find unknown quantities.
  • **Use Consistent Units:** Make sure all measurements are in compatible units (e.g., kilograms for mass, meters per second squared for acceleration).
  • **Solve Step-by-Step:** Start by finding one variable before moving on to others.
In this particular exercise, having already calculated the force, we utilized that calculated force to find how acceleration varies with different mass substitutions. Such a step-by-step approach aids in clear understanding and minimizes errors.

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Most popular questions from this chapter

In moving a 35.0 -kg desk from one side of a classroom to the other, a professor finds that a horizontal force of \(275 \mathrm{~N}\) is necessary to set the desk in motion, and a force of \(195 \mathrm{~N}\) is necessary to keep it in motion at a constant speed. What are the coefficients of (a) static and (b) kinetic friction between the desk and the floor?

(a) When an object is on an inclined plane, the normal force exerted by the inclined plane on the object is (1) less than, (2) equal to, (3) more than the weight of the object. Why? (b) For a \(10-\mathrm{kg}\) object on a \(30^{\circ}\) inclined plane, what are the object's weight and the normal force exerted on the object by the inclined place?

A physicist's car has a small lead weight suspended from a string attached to the interior ceiling. Starting from rest, after a fraction of a second the car accelerates at a steady rate for about \(10 \mathrm{~s}\). During that time, the string (with the weight on the end of it) makes a backward (opposite the acceleration) angle of \(15.0^{\circ}\) from the vertical. Determine the car's (and the weight's) acceleration during the 10 -s interval.

IE oo (a) You are told that an object has zero acceleration. Which of the following is true: (1) The object is at rest; (2) the object is moving with constant velocity; (3) either (1) or (2) is possible; or (4) neither 1 nor 2 is possible. (b) Two forces on the object are \(F_{1}=3.6 \mathrm{~N}\) at \(74^{\circ}\) below the \(+x\) -axis and \(F_{2}=3.6 \mathrm{~N}\) at \(34^{\circ}\) above the \(-x\) -axis. Is there a third force on the object? Why or why not? If there is a third force, what is it?

A student weighing \(800 \mathrm{~N}\) crouches on a scale and suddenly springs vertically upward. His roommate notices that the scale reads 900 N momentarily just as he leaves the scale. With what acceleration does he leave the scale?
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