Problem 47 - University Physics with Modern Physics

Question

CALC A force in the \(+x\) -direction with magnitude \(F(x)=18.0 \mathrm{N}-(0.530 \mathrm{N} / \mathrm{m}) x\) is applied to a 6.00 -kg box that is sitting on the horizontal, frictionless surface of a frozen lake. \(F(x)\) is the only horizontal force on the box. If the box is initially at rest at \(x=0,\) what is its speed after it has traveled 14.0 \(\mathrm{m}\) ?

Step-by-Step Solution

Answer

The speed of the box after traveling 14.0 m is approximately 8.16 m/s.

Step 1: Identify the Given Information

We are given that the force equation is \(F(x) = 18.0 \text{ N} - (0.530 \text{ N/m})x\). The mass of the box is 6.00 kg, and the initial position is \(x = 0\) with the initial velocity being 0 since the box is at rest. We need to find the speed after the box has traveled 14.0 m.

Step 2: Apply Work-Energy Theorem

According to the work-energy theorem, the net work done on an object is equal to its change in kinetic energy: \(W = \Delta KE\). The initial kinetic energy is 0 because the box is at rest.

Step 3: Calculate the Work Done by the Force

The work done by a variable force \(F(x)\) over a distance \(x\) is given by the integral \(W = \int_{0}^{14} F(x) \: dx\). Substitute \(F(x) = 18.0 - 0.530x\) into the integral to get \[W = \int_{0}^{14} (18.0 - 0.530x) \, dx\].

Step 4: Evaluate the Integral

Perform the integration: \(\int (18.0 - 0.530x) \, dx = 18.0x - 0.530\frac{x^2}{2}\). Evaluate this from 0 to 14, giving us: \[W = [18.0(14) - 0.265(14)^2] - [18.0(0) - 0.265(0)^2]\].

Step 5: Perform the Arithmetic

Calculate the values from the evaluated integral: \[W = [252 - 0.265(196)] = [252 - 51.94] = 200.06 \text{ J}\]. The work done on the box is 200.06 J.

Step 6: Relate Work to Kinetic Energy

Since \(W = \Delta KE\) and the initial kinetic energy is 0, we know \(200.06 = \frac{1}{2}mv^2\). Substitute the mass \(m = 6.00\) kg and solve for \(v\).

Step 7: Solve for Final Speed

Set up the equation for kinetic energy: \(200.06 = \frac{1}{2}(6.00)v^2\). Simplify to find \(v^2 = \frac{200.06}{3}\).

Step 8: Calculate the Final Speed

Simplify and take the square root to find \(v\): \(v = \sqrt{66.6867}\), which gives \(v \approx 8.16 \text{ m/s}\).

Key Concepts

Variable Force

When dealing with forces in physics, especially in the context of motion, we often encounter situations where the force isn't constant. This is what we call a **variable force**. In this exercise, the force acting on the box isn't always the same; it changes as the position of the box changes. The force is described by the equation:

• \( F(x) = 18.0 \text{ N} - (0.530 \text{ N/m}) x \)

This means that as the box moves further on the horizontal plane, the force decreases linearly. This variable nature of the force is crucial to understand, as it affects how we calculate the work done by or against them.
By comparing this variable force to a constant force, imagine how different calculations become. A constant force is straightforward – simply multiply force by distance. However, with a variable force, we need more advanced tools, like integration, to find the total work done.

Kinetic Energy

Kinetic energy is the energy of motion. A fundamental concept in physics, it is determined by the mass and velocity of the object:

• Kinetic Energy (\( KE \)) = \( \frac{1}{2}mv^2 \)

In our problem, the box starts at rest, so its initial kinetic energy is zero. As the box starts moving due to the force applied, its kinetic energy changes. The work-energy theorem states that the work done on an object equals the change in its kinetic energy:

• Work = \( \Delta KE = KE_{\text{final}} - KE_{\text{initial}} \)

This relationship helps us calculate the final speed once the work done by the force is known.

Integration

Integration is a powerful tool often used in physics to handle variable forces. It allows us to find the total accumulation of a quantity that changes. In our exercise, since the force isn't constant, we use integration to calculate the work done:

• Work = \( \int_{0}^{14} (18.0 - 0.530x) \, dx \)

Integration helps us add up every small piece of work done over the distance, giving us the total work performed by the variable force on the box.
In our example, the integration results in a quadratic function evaluated at the distance boundaries (0 to 14 meters), leading us to find the total work done as 200.06 Joules.
This concept showcases how we can apply mathematics to solve complex physics problems involving change.

Physics Problem Solving

Physics problems often seem complex at first, but they can be simplified by breaking down the problem into manageable steps. In our exercise, we used a step-by-step approach:

• Identify what is given in the problem and what you need to find.
• Apply relevant physics principles, like the work-energy theorem.
• Use integration to calculate work done when dealing with variable forces.
• Relate work to changes in kinetic energy to solve for the desired quantity, like speed.

These steps not only offer a blueprint for solving physics problems but also help you understand the underlying physics concepts. With practice, you can apply these approaches to other problems involving forces, energy, and motion. The key is to remain patient and systematic, letting the process guide you to the solution.