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A certain force gives an object of mass \(m_{1}\) an acceleration of \(12.0 \mathrm{~m} / \mathrm{s}^{2}\) and an object of mass \(m_{2}\) an acceleration of \(3.30 \mathrm{~m} / \mathrm{s}^{2} .\) What acceleration would the force give to an object of \(\operatorname{mass}(\mathrm{a}) m_{2}-m_{1}\) and (b) \(m_{2}+m_{1} ?\)

Short Answer

Expert verified
(a) 4.55 m/s², (b) 2.59 m/s².

Step by step solution

01

Use Newton's Second Law

Newton's Second Law states that force is the product of mass and acceleration, given by the formula: \[ F = m \cdot a \]For mass \( m_1 \), we have \( F = m_1 \cdot 12.0 \mathrm{~m/s}^2 \) and for mass \( m_2 \), \( F = m_2 \cdot 3.30 \mathrm{~m/s}^2 \).
02

Equate the Forces

Since the same force \( F \) is applied in both cases, we can equate the two expressions for \( F \): \[ m_1 \cdot 12.0 = m_2 \cdot 3.30 \]This gives us a relation between \( m_1 \) and \( m_2 \).
03

Solve for the Ratio of Masses

Rearrange the equation from Step 2 to find the ratio \( \frac{m_1}{m_2} \):\[ \frac{m_1}{m_2} = \frac{3.30}{12.0} \approx 0.275 \].
04

Calculate Acceleration for Mass \( m_2 - m_1 \)

The force \( F \) on mass \( m_2 - m_1 \) will be \[ F = (m_2 - m_1) \cdot a \]Substitute \( F = m_2 \cdot 3.30 \) into the equation:\[ m_2 \cdot 3.30 = (m_2 - m_1) \cdot a \]\[ a = \frac{m_2 \cdot 3.30}{m_2 - m_1} \]Using \( m_1 = 0.275 m_2 \) from Step 3, substitute to find \( a \).
05

Calculate Acceleration for Mass \( m_2 + m_1 \)

For \( m_2 + m_1 \), the force \( F \) results in:\[ m_2 \cdot 3.30 = (m_2 + m_1) \cdot a \]\[ a = \frac{m_2 \cdot 3.30}{m_2 + m_1} \]Substitute \( m_1 = 0.275 m_2 \) to solve for \( a \).
06

Compute Values

For part (a), using \( m_1 = 0.275 m_2 \): \[ a = \frac{3.30}{1 - 0.275} \approx \frac{3.30}{0.725} \approx 4.55 \text{ m/s}^2 \] For part (b):\[ a = \frac{3.30}{1 + 0.275} \approx \frac{3.30}{1.275} \approx 2.59 \text{ m/s}^2 \].

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Key Concepts

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

Force
Force is a fundamental concept in physics and is described as any interaction that, when unopposed, changes the motion of an object. The most common formula to calculate force is given by Newton's Second Law of Motion, represented as:
\[ F = m \cdot a \]This formula states that force (\( F \)) is the product of mass (\( m \)) and acceleration (\( a \)). In our problem, we have two objects with different masses experiencing the same force. By using the given accelerations for masses \( m_1 \) and \( m_2 \), each tied to their respective forces, we understand that although their mass differs, the force applied remains constant. Understanding how force interrelates with mass and acceleration is crucial in solving problems related to motion.
Acceleration
Acceleration is the rate of change of velocity of an object. It tells us how fast something is speeding up or slowing down. When a force is applied to an object, it can cause the object to accelerate, depending on its mass. The relationship is mathematically expressed as:
\[ a = \frac{F}{m} \]From our exercise, two accelerations are provided: \( 12.0 \, \mathrm{m/s}^2 \) for mass \( m_1 \) and \( 3.30 \, \mathrm{m/s}^2 \) for mass \( m_2 \). These accelerations help us understand the response of different masses to the same force. Calculating acceleration with varying mass scenarios allows us to analyze different motion outcomes caused by equivalent forces.
Mass Ratio
Mass ratio is a comparison between the masses of two objects and is an essential component in solving physics problems involving multiple objects and forces. In this exercise, we calculated the mass ratio of the two objects with the equation derived from the equal forces applied on both:
\[ \frac{m_1}{m_2} = \frac{3.30}{12.0} \approx 0.275 \]This ratio describes the relative mass of \( m_1 \) compared to \( m_2 \) and allows us to simplify further calculations by expressing one mass in terms of the other. Knowing mass ratios aids in predicting how different mass combinations will react under the same force, which is necessary to solve complex motion problems.
Problem Solving in Physics
Solving problems in physics often requires breaking down the problem into smaller, manageable steps and applying known laws, like Newton's Second Law. Here’s a method to tackle physics problems:
  • Identify the known quantities and what needs to be solved.
  • Apply relevant laws and formulas, like force = mass x acceleration.
  • Set up equations and, where possible, simplify by substitutions (e.g., using mass ratios).
  • Perform algebraic manipulation to isolate the desired variable.
  • Check units and verify if the result is reasonable in the context of the problem.
This approach helps in systematically analyzing and solving complex physics problems, enhancing understanding and accuracy in finding solutions.

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Most popular questions from this chapter

The Zacchini family was renowned for their human-cannonball act in which a family member was shot from a cannon using either elastic bands or compressed air. In one version of the act, Emanuel Zacchini was shot over three Ferris wheels to land in a net at the same height as the open end of the cannon and at a range of \(69 \mathrm{~m} .\) He was propelled inside the barrel for \(5.2 \mathrm{~m}\) and launched at an angle of \(53^{\circ} .\) If his mass was \(85 \mathrm{~kg}\) and he underwent constant acceleration inside the barrel, what was the magnitude of the force propelling him? (Hint: Treat the launch as though it were along a ramp at \(53^{\circ} .\) Neglect air drag.)

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A firefighter who weighs \(712 \mathrm{~N}\) slides down a vertical pole with an acceleration of \(3.00 \mathrm{~m} / \mathrm{s}^{2},\) directed downward. What are the (a) magnitude and (b) direction (up or down) of the vertical force on the firefighter from the pole and the (c) magnitude and (d) direction of the vertical force on the pole from the firefighter?

A lamp hangs vertically from a cord in a descending elevator that decelerates at \(2.4 \mathrm{~m} / \mathrm{s}^{2}\). (a) If the tension in the cord is \(89 \mathrm{~N},\) what is the lamp's mass? (b) What is the cord's tension when the elevator ascends with an upward acceleration of \(2.4 \mathrm{~m} / \mathrm{s}^{2} ?\)

A car that weighs \(1.30 \times 10^{4} \mathrm{~N}\) is initially moving at \(40 \mathrm{~km} / \mathrm{h}\) when the brakes are applied and the car is brought to a stop in \(15 \mathrm{~m} .\) Assuming the force that stops the car is constant, find (a) the magnitude of that force and (b) the time required for the change in speed. If the initial speed is doubled, and the car experiences the same force during the braking, by what factors are (c) the stopping distance and (d) the stopping time multiplied? (There could be a lesson here about the danger of driving at high speeds.)

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