/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 A box rests on a frozen pond, wh... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 9

A box rests on a frozen pond, which serves as a frictionless horizontal surface. If a fisherman applies a horizontal force with magnitude 48.0 \(\mathrm{N}\) to the box and produces an acceleration of magnitude \(3.00 \mathrm{m} / \mathrm{s}^{2},\) what is the mass of the box?

Short Answer

Expert verified
The mass of the box is 16.0 kg.

Step by step solution

01

Identify Known Values

The problem provides the horizontal force applied to the box and the resulting acceleration produced. We know:- Force \( F = 48.0 \text{ N} \)- Acceleration \( a = 3.00 \text{ m/s}^2 \)
02

Recall the Formula for Force

The force applied on an object is calculated using Newton's second law of motion, which is:\( F = m \, a \)where \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration.
03

Rearrange the Formula for Mass

We need to find the mass \( m \), so we rearrange the formula:\( m = \frac{F}{a} \)
04

Substitute Known Values into the Formula

Using the known values, substitute them into the rearranged formula:\( m = \frac{48.0 \text{ N}}{3.00 \text{ m/s}^2} \)
05

Calculate the Mass

Perform the division to find the mass:\( m = 16.0 \text{ kg} \)
06

Verify Units

The units for mass are in kilograms (kg), which is appropriate as Force (N) divided by Acceleration (m/s虏) results in kg.

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Key Concepts

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

Force and Acceleration
In the context of Newton's second law of motion, force and acceleration have a direct relationship. Newton's second law states that the force exerted on an object is proportional to the acceleration of that object multiplied by its mass, which is mathematically represented as:\[ F = m \cdot a \]- **Force (F)**: Measured in newtons (N), it is the push or pull applied to an object. In our exercise, a force of 48.0 N is applied.- **Acceleration (a)**: This is the rate at which an object's velocity changes. It is measured in meters per second squared (m/s虏). Here, the acceleration is 3.00 m/s虏.The relationship indicates that for a constant mass, any increase in force will result in an increase in acceleration. Conversely, for a given force, increasing the mass will decrease the acceleration. This intuitive interaction is foundational for understanding how objects move in response to applied forces.
Frictionless Surface
A frictionless surface is an idealized concept, often used in physics problems to simplify calculations. In reality, most surfaces will have some degree of friction that opposes motion. However, describing the surface as frictionless allows us to ignore frictional forces that would otherwise complicate the calculation. - **No Opposing Forces**: Since the surface is frictionless, there are no forces opposing the movement of the box, aside from the force applied. - **Ideal Conditions**: This means that all the force applied to the box is used solely for acceleration, making it easier to apply Newton's second law without adjustments for friction. By removing friction from the equation, the problem becomes a straightforward application of the force to achieve the acceleration. This assumption helps break down the larger concepts in a physics problem into simpler, more manageable components.
Mass Calculation
Calculating mass in problems like this one involves rearranging Newton's second law. Here, we are given both the force and acceleration, and we need to find the mass of the object.The formula by Newton's second law is \( F = m \cdot a \). By rearranging this formula, we solve for mass \( m \):\[ m = \frac{F}{a} \]- **Substitution of Values**: Substitute the given values: force (48.0 N) and acceleration (3.00 m/s虏).Using these values, calculate the mass of the box:\[ m = \frac{48.0 \text{ N}}{3.00 \text{ m/s}^2} = 16.0 \text{ kg} \]- **Verification of Units**: This calculation confirms that the units are consistent with mass being measured in kilograms (kg) since newtons divided by meters per second squared result in kilograms.This approach ensures accuracy in determining the mass and helps students practice manipulating equations to solve for a desired variable.

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Most popular questions from this chapter

A dockworker applies a constant horizontal force of 80.0 \(\mathrm{N}\) to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves 11.0 \(\mathrm{m}\) in 5.00 \(\mathrm{s}\) . (a) What is the mass of the block of ice? (b) If the worker stops pushing at the end of 5.00 \(\mathrm{s}\) s, how far does the block move in the next 5.00 \(\mathrm{s?}\)

To study damage to aircraft that collide with large birds, you design a test gun that will accelerate chicken-sized objects so that their displacement along the gun barrel is given by \(x=\) \(\left(9.0 \times 10^{3} \mathrm{m} / \mathrm{s}^{2}\right) t^{2}-\left(8.0 \times 10^{4} \mathrm{m} / \mathrm{s}^{3}\right) t^{3} .\) The object leaves the end of the barrel at \(t=0.025 \mathrm{s}\) (a) How long must the gun barrel be? \((b)\) What will be the speed of the objects as they leave the end of the barrel? (c) What net force must be exerted on a \(1.50-k g\) object at \((\mathrm{i}) t=0\) and (ii) \(t=0.025 \mathrm{s} ?\)

A loaded elevator with very worn cables has a total mass of 2200 \(\mathrm{kg}\) , and the cables can withstand a maximum tension of \(28,000 \mathrm{N} .\) (a) Draw the free-body force diagram for the elevator. In terms of the forces on your diagram, what is the net force on the elevator? Apply Newton's second law to the elevator and find the maximum upward acceleration for the elevator if the cables are not to break. (b) What would be the answer to part (a) if the elevator were on the moon, where \(g=1.62 \mathrm{m} / \mathrm{s}^{2} ?\)

If we know \(F(t),\) the force as a function of time, for straight-line motion, Newton's second law gives us \(a(t),\) the acceleration as a function of time. We can then integrate \(a(t)\) to find \(v(t)\) and \(x(t)\) . However, suppose we know \(F(v)\) instead. (a) The net force on a body moving along the \(x\) -axis equals \(-C v^{2} .\) Use Newton's second law written as \(\Sigma F=m d v / d t\) and two integrations to show that \(x-x_{0}=(m / C) \ln \left(v_{0} / v\right) .\) (b) Show that Newton's second law can be written as \(\Sigma F=m v d v / d x .\) Derive the same expression as in part (a) using this form of the second law and one integration.

A ball is hanging from a long string that is tied to the ceiling of a train car traveling castward on horizontal tracks. An observer inside the train car sees the ball hang motionless. Draw a clearly labeled free-body diagram for the ball if (a) the train has a uniform velocity, and (b) the train is speeding up uniformly. Is the net force on the ball zero in either case? Explain.
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