Problem 60 - University Physics with Modern Physics

Question

CALC A balky cow is leaving the barn as you try harder and harder to push her back in. In coordinates with the origin at the barn door, the cow walks from \(x=0\) to \(x=6.9 \mathrm{m}\) as you apply a force with \(x\) -component \(F_{x}=-[20.0 \mathrm{N}+(3.0 \mathrm{N} / \mathrm{m}) x] .\) How much work does the force you apply do on the cow during this displacement?

Step-by-Step Solution

Answer

The work done is \(-209.505 \text{ J}\).

Step 1: Understand the Work Done Formula

Work done by a force is given by the integral of the force over the displacement. Mathematically, this is expressed as \( W = \int F_x \, dx \). In this case, the force \( F_x \) depends on the position \( x \).

Step 2: Set Up the Integral

Substitute the given force expression \( F_x = -[20.0 + 3.0x] \) into the work formula. The work done becomes \( W = \int_{0}^{6.9} -[20.0 + 3.0x] \, dx \).

Step 3: Calculate the Integral

Evaluate \( \int_{0}^{6.9} -[20.0 + 3.0x] \, dx \). This is done by integrating each term separately:1. \( -\int 20.0 \, dx = -20.0x \)2. \( -\int 3.0x \, dx = -1.5x^2 \) (since \( \int x \, dx = \frac{1}{2}x^2 \))Thus, \( W = [-20.0x - 1.5x^2]_{0}^{6.9} \).

Step 4: Evaluate the Integral at the Limits

Plug in the limits to evaluate:- Upper limit: \( [-20.0(6.9) - 1.5(6.9)^2] \)- Lower limit: \( [-20.0(0) - 1.5(0)^2] = 0 \)Calculate the upper limit value to find:\( -20.0 \times 6.9 = -138.0 \)\( -1.5 \times (6.9)^2 = -71.505 \)Sum these to get: \( -138.0 - 71.505 = -209.505 \).

Step 5: Interpret the Result

The work done by the force is \(-209.505 \text{ J}\). The negative sign indicates that the force applied does work against the cow's movement from \(x = 0\) to \(x = 6.9 \) meters.

Key Concepts

Integral Calculus

Integral calculus plays a crucial role in understanding how to compute work done by varying forces. At its core, integral calculus allows us to sum up tiny pieces of work done over small distances to find the total work. For instance, when a force varies with position, calculus lets us integrate the force function over a specified distance interval.
This concept is often challenging to grasp, but it can be visualized as finding the area under the curve of a force-position graph. The integral can be thought of as an accumulation of work done over every tiny segment of displacement.
In this example, we integrated the force function: \( F_x = -[20.0 + 3.0x] \), from \(x = 0\) to \(x = 6.9\) meters.

• The integral expressed mathematically is \( W = \int_{0}^{6.9} -[20.0 + 3.0x] \, dx \).
• Each tiny calculation of work is summed into this integral between the start and stop points, giving the total work done.

Force and Displacement

Force and displacement are core concepts in physics that interrelate when calculating work done. Work is defined as the product of displacement and the component of the force in the direction of the displacement.
When the force is constant, this is simply \( W = F \cdot d \). But in scenarios where force changes over distance, or isn't entirely in the direction of displacement, calculus comes into play.

• The given force expression \( F_x = -[20.0 + 3.0x] \) means the force applied changes as the cow moves further.
• The position dependence means the force is stronger in one direction and must be integrated over the path the cow takes.

The displacement, \(6.9 \) meters in this case, is crucial because it defines the interval over which we calculate the integral. Understanding the movement from start to finish allows accurate calculation of work done by exploring all changing forces applied along the path.

Physics Problem-Solving

Solving physics problems, especially those involving work done by varying forces, requires a structured approach. Here is a simple strategy which can help:
First, understand the problem context, like knowing that work is involved with a changing force applied over a specific distance. This brings us to the physics concepts involved: force, displacement, and the work formula.

• Break down the force function to see how it varies with position. Use the given expression and apply it to the formula for work.
• Set up the integral to calculate work done over the specified displacement. This turns the qualitative problem into a quantitative one.
• Solve the integral, which typically involves evaluating the integral at designated limits.

In this problem, solving it step-by-step entailed integrating \(-[20.0 + 3.0x]\), checking calculations at limits of \(x=0\) to \(x=6.9\), and yielding the negative work value which indicated the force applied opposes the cow’s movement.