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Chapter 3: Problem 14

A \(600-\mathrm{N}\) object is to be given an acceleration of \(0.70 \mathrm{~m} / \mathrm{s}^{2} .\) How large an unbalanced force must act upon it? Notice that the weight, not the mass, of the object is given. Assuming the weight was measured on the Earth, use \(F_{W}=m g\) to find $$ m=\frac{F_{W}}{g}=\frac{600 \mathrm{~N}}{9.81 \mathrm{~m} / \mathrm{s}^{2}}=61.2 \mathrm{~kg} $$ Now that we know the mass of the object \((61.2 \mathrm{~kg})\) and the desired acceleration \(\left(0.70 \mathrm{~m} / \mathrm{s}^{2}\right)\), the force is $$ F=m a=(61.2 \mathrm{~kg})\left(0.70 \mathrm{~m} / \mathrm{s}^{2}\right)=42.8 \mathrm{~N} \text { or } 43 \mathrm{~N} $$

Short Answer

Expert verified
The unbalanced force required is 43 N.

Step by step solution

01

Identify Given Information

We are given the weight of the object as 600 N and the desired acceleration as 0.70 m/s虏. The weight was measured on Earth, implying the use of Earth's gravity.
02

Calculate Mass from Weight

To find the mass, use the formula: \( F_W = m \cdot g \). Rearrange to get \( m = \frac{F_W}{g} \), where \( F_W = 600 \text{ N} \) and \( g = 9.81 \text{ m/s}^2 \). This gives: \[ m = \frac{600 \text{ N}}{9.81 \text{ m/s}^2} = 61.2 \text{ kg} \].
03

Apply Newton's Second Law

Using the formula \( F = m \cdot a \), where \( m = 61.2 \text{ kg} \) is the mass and \( a = 0.70 \text{ m/s}^2 \) is the acceleration, calculate the force: \[ F = (61.2 \text{ kg}) \times (0.70 \text{ m/s}^2) \].
04

Calculate the Unbalanced Force

From the previous step calculation, \( F = (61.2 \text{ kg}) \times (0.70 \text{ m/s}^2) = 42.84 \text{ N} \). Rounding to the nearest whole number, the answer is 43 N.

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Key Concepts

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

Unbalanced Force
An unbalanced force occurs when the total forces acting on an object do not cancel each other out. This means that there is a net force, which results in the object鈥檚 motion changing in some way, like accelerating or decelerating. For instance, if two forces push against each other with unequal strength, the stronger force will dominate, causing the object to move.
Newton's Second Law of Motion states that the force acting on an object is equal to the mass of that object multiplied by its acceleration, formulated as \(F = ma\).
This law helps us determine how much force is needed to move an object or alter its motion based on its mass and the desired acceleration.
In situations where forces are not balanced, we see a change in speed or direction, indicating the presence of an unbalanced force.
Calculating Mass from Weight
Weight and mass are related, but they are not the same. Weight is the force of gravity acting on an object, and it can vary based on the object's location (like being on the moon vs. being on Earth).
Mass, however, is a measure of the amount of matter in an object and does not change with location. To find mass when you know the weight, you can use the equation \(F_W = mg\), where:
  • \(F_W\) is the weight (in Newtons)
  • \(m\) is the mass (in kilograms)
  • \(g\) is the acceleration due to gravity (on Earth, \(g = 9.81 \, \text{m/s}^2\))
To calculate mass, rearrange the formula to \(m = \frac{F_W}{g}\). So, if a 600-N object is measured on Earth, its mass would be \(m = \frac{600 \, \text{N}}{9.81 \, \text{m/s}^2} = 61.2 \, \text{kg}\).
Understanding this relationship helps you determine an object's mass from its weight across different scenarios.
Acceleration Calculation
Acceleration refers to the rate at which an object's speed or direction changes. According to Newton's Second Law, the acceleration of an object depends on both the net force acting upon the object and the mass of the object.
Using the formula \(F = ma\), you can calculate acceleration as \(a = \frac{F}{m}\), where:
  • \(F\) is the net force applied to the object
  • \(m\) is the mass of the object
  • \(a\) is the resulting acceleration
In the specific scenario where a mass of 61.2 kg is subjected to an overall force of 43 N, we can confirm the calculated acceleration: \(a = \frac{43 \, \text{N}}{61.2 \, \text{kg}} = 0.70 \, \text{m/s}^2\).
Accelerating an object requires an understanding of how mass and force interact to produce movement.

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Most popular questions from this chapter

An object has a mass of \(300 \mathrm{~g}\). (a) What is its weight on Earth? ( \(b\) ) What is its mass on the Moon? ( \(c\) ) What will be its acceleration on the Moon under the action of a \(0.500-\mathrm{N}\) resultant force?

A force of \(100 \mathrm{~N}\) makes an angle of \(\theta\) with the \(x\) -axis and has a scalar \(y\) -component of \(30 \mathrm{~N}\). Find both the scalar \(x\) -component of the force and the angle \(\theta\). (Remember that the number \(100 \mathrm{~N}\) has three significant figures whereas \(30 \mathrm{~N}\) has only two.)Begin your analysis by drawing a diagram. Here the data are sketched roughly in Fig. \(3-4\). We wish to find \(F_{x}\) and \(\theta\). Since $$ \sin \theta=\frac{30 \mathrm{~N}}{100 \mathrm{~N}}=0.30 $$ \(\theta=17.46^{\circ}\), and thus, to two significant figures, \(\theta=17^{\circ} .\) Then, using the \(\cos \theta\) $$ F_{x}=(100 \mathrm{~N}) \cos 17.46^{\circ}=95 \mathrm{~N} $$

A 300 -g mass hangs at the end of a string. A second string hangs from the bottom of that mass and supports a \(900-\mathrm{g}\) mass. ( \(a\) ) Find the tension in each string when the masses are accelerating upward at \(0.700 \mathrm{~m} / \mathrm{s}^{2}\). Don't forget gravity. (b) Find the tension in each string when the acceleration is \(0.700 \mathrm{~m} / \mathrm{s}^{2}\) downward.

An elevator starts from rest with a constant upward acceleration. It moves \(2.0 \mathrm{~m}\) in the first \(0.60 \mathrm{~s}\). A passenger in the elevator is holding a \(3.0-\mathrm{kg}\) package by a vertical string. What is the tension in the string during the accelerating process?

Consider an essentially spherical homogeneous celestial body of mass \(M\). The acceleration due to gravity in its vicinity beyond its surface at a distance \(R\) from its center is \(g_{R}\). Show that $$ g_{R}=\frac{G M}{R^{2}} $$ Notice that the acceleration drops off as \(1 / R^{2}\). Imagine an object of mass \(m\) at a distance \(R\) from the center of our celestial body. Its weight is \(F_{W}=m g_{R}\), but that's also the gravitation force on it due to the mass \(M\), that is, \(F_{W}=F_{G}\). Hence, $$ m g_{R}=G \frac{m M}{R^{2}} $$
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