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Chapter 4: Problem 3

A \(6.0-\mathrm{kg}\) object undergoes an acceleration of \(2.0 \mathrm{~m} / \mathrm{s}^{2}\). (a) What is the magnitude of the resultant force acting on it? (b) If this same force is applied to a \(4.0\) -kg object, what acceleration is produced?

Short Answer

Expert verified
a) The magnitude of the resultant force on the 6.0 kg object is 12 N. b) The acceleration caused by the same force acting on a 4.0 kg object is 3.0 m/s虏.

Step by step solution

01

Calculate the force on the 6.0 kg object

According to Newton鈥檚 second law, force is the product of mass and acceleration. This can be formulated as Force = Mass x Acceleration. Here, the object's mass \(M_1\) is 6 kg and its acceleration \(A_1\) is 2 m/s虏. So, the force \(F\) acting on it can be calculated by multiplying the mass and acceleration, \(F = M_1 * A_1 \).
02

Compute the value

Lets put the values into the formula: \(F = 6.0 \, kg * 2.0 \, m/s^{2}\) which equals to \(12 \, N\) (Newton).
03

Determine the acceleration of the 4.0 kg object under the same force

We know that \(F = M_2 * A_2 \) where \(M_2\) is the mass of the second body and \(A_2\) is its acceleration. Rearranging the formula for acceleration we have \( A_2 = F / M_2\)
04

Calculating the new acceleration

Inserting our values into the formula gives us: \(A_2 = 12 \, N / 4.0 \, kg\) which equals to \(3.0 \, m/s^{2}\)

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Key Concepts

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

Force Calculation
Understanding force calculation is essential for solving physics problems related to Newton's second law, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. This fundamental relationship can be expressed with the equation:

\( F = m \times a \)
Where F represents the force in newtons (N), m the mass in kilograms (kg), and a the acceleration in meters per second squared (m/s虏). When you're faced with a problem where you need to calculate the force, start by identifying the mass and acceleration. Then, simply multiply these two values to find the force.

Applying the Formula

For example, in the given exercise, to calculate the force acting on a 6.0-kg object accelerating at 2.0 m/s虏, we multiply 6.0 kg (mass) by 2.0 m/s虏 (acceleration) to get 12 N (force).
Mass and Acceleration
The concepts of mass and acceleration are intertwined in physics, particularly when dealing with Newton's second law. Mass is a measure of how much matter is in an object and is often the more straightforward value, as it remains constant regardless of location. Acceleration, on the other hand, is the rate at which the velocity of an object changes over time, which can be influenced by various forces.

Understanding Acceleration

When solving problems, you'll often calculate acceleration by dividing the force by the mass of the object. This is derived from rearranging the equation of Newton's second law to:

\( a = \frac{F}{m} \)
If a force is applied to two objects of different masses, the object with less mass will experience a greater acceleration. This is why, in the exercise, applying the same force to a lighter object (4.0 kg) results in a higher acceleration (3.0 m/s虏) than the heavier object (6.0 kg).
Unit Conversion
Unit conversion is a crucial step in physics calculations, allowing different measurements to be compared and utilized in formulas correctly. In the context of force calculation, we primarily deal with kilograms (kg) for mass, meters per second squared (m/s虏) for acceleration, and newtons (N) for force.Understanding the Newton
A newton is a derived SI unit of force. It is defined as the amount of force required to accelerate a one-kilogram mass by one meter per second squared. Therefore, when you multiply mass in kilograms by acceleration in meters per second squared, the result is expressed in newtons.When solving physics problems, ensuring that all the units match up before performing calculations is crucial. It's common to convert grams to kilograms or centimeters per second squared to meters per second squared to maintain consistency within the SI unit system. Proper unit conversion prevents errors and ensures accurate results in your calculations.

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Most popular questions from this chapter

A magician pulls a tablecloth from under a \(200-\mathrm{g} \mathrm{mug}\) located \(30.0 \mathrm{~cm}\) from the edge of the cloth. The cloth exerts a friction force of \(0.100 \mathrm{~N}\) on the mug and is pulled with a constant acceleration of \(3.00 \mathrm{~m} / \mathrm{s}^{2}\). How far does the mug move relative to the horizontal tabletop before the cloth is completely out from under it? Note that the cloth must move more than \(30 \mathrm{~cm}\) relative to the tabletop during the process.

An \(80-\mathrm{kg}\) stuntman jumps from a window of a building situated \(30 \mathrm{~m}\) above a catching net. Assuming air resistance exerts a \(100-\mathrm{N}\) force on the stuntman as he falls, determine his velocity just before he hits the net.

(a) An clevator of mass \(m\) moving upward has two forces acting on it: the upward force of tension in the cable and the downward force due to gravity. When the elevator is accelerating upward, which is greater, \(T\) or \(w ?^{2}\) (b) When the elevator is moving at a constant velocity upward, which is greater, \(T\) or \(w ?\) (c) When the elevator is moving upward, but the acceleration is downward, which is greater, Tor w? (d) Let the elevator have a mass of \(1500 \mathrm{~kg}\) and an upward acceleration of \(2.5 \mathrm{~m} / \mathrm{s}^{2}\), Find \(T\). Is your answer consistent with the answer to part (a)? (e) The elevator of part (d) now moves with a constant upward velocity of \(10 \mathrm{~m} / \mathrm{s}\). Find \(T\). Is your answer consistent with your answer to part (b)? (f) Having initially moved upward with a constant velocity, the elevator begins to accelerate downward at \(1.50 \mathrm{~m} / \mathrm{s}^{2}\), Find \(T\). Is your answer consistent with your answer to part (c)?

An object of mass \(m\) is dropped from the roof of a building of height \(h .\) While the object is falling, a wind blowing parallel to the face of the building exerts a constant horizontal force \(F\) on the object. (a) How long does it take the object to strike the ground? Express the time \(t\) in terms of \(g\) and \(h\) (b) Find an expression in terms of \(m\) and \(F\) for the acceleration \(a_{x}\) of the object in the horizontal direction (taken as the positive \(x\) -direction). (c) How far is the object displaced horizontally before hitting the ground? Answer in terms of \(m, g, F_{t}\) and \(h\). \((d)\) Find the magnitude of the object's acceleration while it is falling, using the variables \(F, m\), and \(g\).

Objects of masses \(m_{1}=4.00 \mathrm{~kg}\) and \(m_{2}=9.00 \mathrm{~kg}\) are connected by a light string that passes over a frictionless pulley as in Figure \(\mathrm{P} 4.54\). The object \(m_{1}\) is held at rest on the floor, and \(m_{2}\) rests on a fixed incline of \(\theta=40.0^{\circ}\). The objects are released from rest, and \(m_{2}\) slides \(1.00 \mathrm{~m}\) down the incline in \(4.00 \mathrm{~s}\). Determine (a) the acceleration of each object, (b) the tension in the string, and (c) the coefficient of kinetic friction between \(m_{2}\) and the incline.
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