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Chapter 7: Problem 57

ssm One average force \(\overrightarrow{\mathbf{F}}_{1}\) has a magnitude that is three times as large as that of another average force \(\overrightarrow{\mathbf{F}}_{2}\) . Both forces produce the same impulse. The average force \(\overrightarrow{\mathbf{F}}_{1}\) acts for a time interval of 3.2 \(\mathrm{ms}\) . For what time interval does the average force \(\overline{\mathbf{F}}_{2}\) act?

Short Answer

Expert verified
The time interval \( \Delta t_2 \) is 9.6 ms.

Step by step solution

01

Understanding Impulse

Impulse is given by the product of the average force and the time duration for which the force acts. The formula is: \( J = F \cdot \Delta t \). Given that both forces produce the same impulse, we have the equation \( F_1 \cdot \Delta t_1 = F_2 \cdot \Delta t_2 \).
02

Relate the Forces

From the problem, we know that the magnitude of \( F_1 \) is three times that of \( F_2 \). So, \( F_1 = 3F_2 \).
03

Substitute and Solve for \( \Delta t_2 \)

Substitute \( F_1 = 3F_2 \) and \( \Delta t_1 = 3.2 \) ms into the impulse equation. We obtain: \( 3F_2 \cdot 3.2 = F_2 \cdot \Delta t_2 \). Cancel \( F_2 \) from both sides to get \( 3 \times 3.2 = \Delta t_2 \).
04

Calculate \( \Delta t_2 \)

Now, calculate \( \Delta t_2 \) by performing the multiplication: \( \Delta t_2 = 9.6 \) ms.

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Key Concepts

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

Average Force
In physics, when we speak about average force, we are looking at the consistent force applied over a certain time period. It is calculated by dividing the change in momentum by the time interval during which the change occurs. The average force provides useful insights into real-world scenarios, like understanding how long a force needs to act to change the motion of an object.
In the given problem, force \(F_1\) is three times stronger than force \(F_2\). Yet, both forces generate identical impulses. Thus, average force plays a crucial role in balancing these differences through time intervals.
Time Interval
Time interval is the duration for which a force is applied. In the context of impulse, it's a core factor that, alongside force, determines the total impulse. The command over time intervals helps in controlling the outcome effect of forces.
In the exercise, while \(F_1\) acts for 3.2 milliseconds, we computed that \(F_2\) had to act for a longer 9.6 milliseconds to produce the same impulse. This demonstrates how a smaller force must act over a longer time to achieve the same effect as a larger force in a shorter time.
Impulse Equation
The impulse equation is fundamental in linking force, time, and movement. It is given by: \( J = F \cdot \Delta t \). This equation tells us that impulse \(J\), which alters the momentum of an object, equals force \(F\) multiplied by the time interval \(\Delta t\) for which the force is applied.
In this context, both forces deliver the same impulse, \(J\). Hence, despite \(F_1\) being significantly stronger than \(F_2\), the longer time interval for \(F_2\) balances the equation, reinforcing impulse's dependence on both force and time.
Physics Problem Solving
Solving physics problems involves both understanding theoretical concepts and applying them practically. The solution requires translating the problem鈥檚 conditions into equations and solving them logically.
Here, we broke down the question by understanding what was asked about average force and time intervals. By employing the impulse equation and known relationships like \(F_1 = 3F_2\) and \(\Delta t_1 = 3.2 \text{ ms}\), we derived \(\Delta t_2\). Problem-solving in physics often necessitates careful calculation and logical deduction to find the solution.

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Most popular questions from this chapter

ssm A lumberjack (mass \(=98 \mathrm{kg} )\) is standing at rest on one end of a floating log (mass \(=230 \mathrm{kg}\) ) that is also at rest. The lumberjack runs to the other end of the log, attaining a velocity of \(+3.6 \mathrm{m} / \mathrm{s}\) relative to the shore, and then hops onto an identical floating log that is initially at rest. Neglect any friction and resistance between the logs and the water. (a) What is the velocity of the first log just before the lumberjack jumps off? (b) Determine the velocity of the second log if the lumberjack comes to rest on it.

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ssm A projectile (mass \(=0.20 \mathrm{kg}\) is fired at and embeds itself in a stationary target (mass \(=2.50 \mathrm{kg}\) . With what percentage of the projectile's incident kinetic energy does the target (with the projectile in it) fly off after being struck?
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