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An unknown force acting on a \(50.0\) -g body floating in space produces a constant acceleration of \(20.0 \mathrm{~cm} / \mathrm{s}^{2}\). If the same force is now made to act on a different body, also in space, producing a constant acceleration of \(40.0 \mathrm{~cm} / \mathrm{s}^{2}\), what is the mass of that body?

Short Answer

Expert verified
The mass of the second body is 25.0 g.

Step by step solution

01

Identify Variables and Formula

Let's identify the given variables first. We know the mass of the first object \( m_1 = 50.0 \text{ g} = 0.050 \text{ kg} \) (converted to kg) and its acceleration \( a_1 = 20.0 \text{ cm/s}^2 = 0.20 \text{ m/s}^2 \) (converted to m/s²). For the second body, the acceleration \( a_2 = 40.0 \text{ cm/s}^2 = 0.40 \text{ m/s}^2 \). We need to find its mass \( m_2 \). The force equation is \( F = m \times a \).
02

Calculate the Force Acting on the First Body

Using the formula \( F = m \times a \), we calculate the force exerted on the first body. Substitute \( m_1 = 0.050 \text{ kg} \) and \( a_1 = 0.20 \text{ m/s}^2 \):\[F = 0.050 \times 0.20 = 0.010 \text{ N}\]This force is \(0.010 \text{ N}\).
03

Use Same Force on Second Body to Find Mass

Now, the same force \( F = 0.010 \text{ N} \) is used on the second body. We know the acceleration \( a_2 = 0.40 \text{ m/s}^2 \). Use \( F = m_2 \times a_2 \) to find \( m_2 \):\[m_2 = \frac{F}{a_2} = \frac{0.010}{0.40} = 0.025 \text{ kg}\]So, the mass of the second body is \(0.025 \text{ kg}\) or \(25.0 \text{ g}\).

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Key Concepts

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

Force and Acceleration
Newton's Second Law is fundamental in understanding the relationship between force, mass, and acceleration. It is succinctly expressed with the formula: \[F = m \times a\],where \(F\) stands for force, \(m\) for mass, and \(a\) for acceleration.

This law reveals that the force exerted on an object is directly proportional to both its mass and the acceleration it experiences. If you have a constant force applied to two different objects, like in our problem, the acceleration will vary inversely with the mass of the object.

Therefore, a smaller mass will experience a larger acceleration for the same applied force. Grasping this concept helps you understand how changing one variable affects the others:
  • Increased mass leads to decreased acceleration if the force remains constant.
  • Decreased mass results in increased acceleration under the same force.
  • Proportionality is key: if you double the force and keep the mass the same, the acceleration doubles.
Understanding these principles allows you to predict physical behavior in a variety of scenarios.
Mass Calculation
Calculating the mass of an object based on the given force and acceleration is straightforward when using Newton's Second Law. Let's delve into the process.

In our example, the first step was to determine the force exerted on the first body by using the acceleration \(a_1\) and the known mass \(m_1\). By substituting into the equation \(F = m_1 \times a_1\), we find the force of \(0.010 \text{ N}\).

With this known force, we apply it to the second object, using its given acceleration \(a_2\), to find its mass \(m_2\). In mathematical terms, we rearrange the formula to solve for mass: \[m_2 = \frac{F}{a_2}\].

By substituting \(F = 0.010 \text{ N}\) and \(a_2 = 0.40 \text{ m/s}^2\), we find that \(m_2 = 0.025 \text{ kg}\). Knowing how to rearrange and solve the equation allows you to compute any missing variable when given two others.
Physics Problem Solving
Solving physics problems systematically is vital for success. Breaking down each problem into parts focuses your approach and helps ensure accuracy.

Here’s a structured approach:
  • **Identify the Variables:** Clearly list all the known values, such as mass, force, and acceleration, as well as the unknown variables you need to find.
  • **Convert Units:** Make sure all units are consistent. For example, convert grams to kilograms and centimeters per second squared to meters per second squared depending on the context.
  • **Formulate the Equation:** Use the appropriate equation for the situation, like Newton's Second Law in our example.
  • **Complete the Calculations:** Perform the necessary calculations while paying close attention to each step to avoid errors.

Each problem you solve using this methodical approach builds your understanding and helps develop strong problem-solving skills. Whether in exams or real-world applications, these skills are invaluable.

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Most popular questions from this chapter

A car whose weight is \(F_{W}\) is on a ramp, which makes an angle \(\theta\) to the horizontal. How large a perpendicular force must the ramp withstand if it is not to break under the car's weight? As rendered in Fig. 3-6, the car's weight is a force \(\overrightarrow{\mathrm{F}}_{W}\) that pulls straight down on the car. We take components of \(\overrightarrow{\mathrm{F}}\) along the incline and perpendicular to it. The ramp must balance the force component \(F_{W} \cos \theta\) if the car is not to crash through the ramp. In other words, the force exerted on the car by the ramp, upwardly perpendicular to the ramp, is \(F_{N}\) and \(F_{N}=F_{W} \cos \theta\).

A child pulls on a rope attached to a sled with a force of \(60 \mathrm{~N}\). The rope makes an angle of \(40^{\circ}\) to the ground. (a) Compute the effective value of the pull tending to move the sled along the ground. (b) Compute the force tending to lift the sled vertically. As depicted in Fig. \(3-5\), the components of the \(60 \mathrm{~N}\) force are \(39 \mathrm{~N}\) and \(46 \mathrm{~N}\). ( \(a\) ) The pull along the ground is the horizontal component, 46 N. (b) The lifting force is the vertical component, \(39 \mathrm{~N}\).

A 600 -kg car is coasting along a level road at \(30 \mathrm{~m} / \mathrm{s}\). (a) How large a retarding force (assumed constant) is required to stop it in a distance of \(70 \mathrm{~m} ?(b)\) What is the minimum coefficient of friction between tires and roadway if this is to be possible? Assume the wheels are not locked, in which case we are dealing with static friction- there's no sliding. ( \(a\) ) First find the car's acceleration from a constant- \(a\) equation. It is known that \(\mathrm{v}_{i x}=30 \mathrm{v}_{f x},=0\), and \(x=70 \mathrm{~m}\). Use \(v_{k}^{2}=v_{u}^{2}+2 a x\) to find $$ a=\frac{v_{f x}^{2}-v_{i r}^{2}}{2 x}=\frac{0-900 \mathrm{~m}^{2} / \mathrm{s}^{2}}{140 \mathrm{~m}}=-6.43 \mathrm{~m} / \mathrm{s}^{2} $$ Now write $$ F=m a=(600 \mathrm{~kg})\left(-6.43 \mathrm{~m} / \mathrm{s}^{2}\right)-3858 \mathrm{~N} \text { or }-3.9 \mathrm{kN} $$ (b) Assume the force found in \((a)\) is supplied as the friction force between the tires and roadway. Therefore, the magnitude of the friction force on the tires is \(F_{\mathrm{f}}=3858 \mathrm{~N}\). The coefficient of friction is given by \(u=F_{f} / F_{N}\), where \(F_{N}\) is the normal force. In the present case, the roadway pushes up on the car with a force equal to the car's weight. Therefore, that $$ \begin{array}{l} F_{N}=F_{W}=m g=(600 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=5886 \mathrm{~N} \\ \mu_{s}=\frac{F_{\mathrm{f}}}{F_{N}}=\frac{3858}{5886}=0.66 \end{array} $$ The coefficient of friction must be at least \(0.66\) if the car is to stop within \(70 \mathrm{~m}\).

Typically, a bullet leaves a standard 45-caliber pistol (5.0-in. barrel) at a speed of \(262 \mathrm{~m} / \mathrm{s}\). If it takes \(1 \mathrm{~ms}\) to traverse the barrel, determine the average acceleration experienced by the \(16.2\) -g bullet within the gun, and then compute the average force exerted on it.

Having hauled it to the top of a tilted driveway, a child is holding a wagon from rolling back down. The driveway is inclined at \(20^{\circ}\) to the horizontal. If the wagon weighs \(150 \mathrm{~N}\), with what force must the child pull on the handle if the handle is parallel to and pointing up the incline?

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