Question

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 \(x = 6.9\) m as you apply a force with \(x\)-component \(F_x = - [20.0 \, \mathrm{N} + (3.0 \, \mathrm{N/m})x]\). How much work does the force you apply do on the cow during this displacement?

Short answer

The work done on the cow is \(-209.415 \, \text{J}\).

Step-by-step solution

Recall the Work Formula

The work done by a variable force along a straight path is given by the integral of the force over the displacement. Mathematically, this is represented as \( W = \int F_x \, dx \). Here, \( F_x \) is the force component in the direction of displacement, which in this problem is given as a function \( F_x = - [20.0 \, \mathrm{N} + (3.0 \, \mathrm{N/m})x] \).

Set Up the Integral

We need to integrate the given force function from the initial position \( x = 0 \) to the final position \( x = 6.9 \) meters. Therefore, set up the integral as:\[ W = \int_{0}^{6.9} -[20.0 + 3.0x] \, dx \]

Solve the Integral

Find the indefinite integral first:\[ \int -[20 + 3x] \, dx = -20x - \frac{3}{2}x^2 + C \]where \( C \) is the integration constant. We then evaluate this from 0 to 6.9:\[ W = \left(-20x - \frac{3}{2}x^2\right) \bigg|_0^{6.9} \]

Plug in the Limits

Plug in \( x = 6.9 \) and \( x = 0 \) into the expression:\[ W = \left( -20(6.9) - \frac{3}{2}(6.9)^2 \right) - \left( -20(0) - \frac{3}{2}(0)^2 \right) \]

Calculate Numerically

Perform the calculations:- For \( x = 6.9 \): \[ -20 \times 6.9 = -138.0 \] \[ -\frac{3}{2} \times 6.9^2 = -\frac{3}{2} \times 47.61 = -71.415 \]- For \( x = 0 \): \[ -20 \times 0 - \frac{3}{2} \times 0^2 = 0 \]Subtract the two results:\[ W = (-138.0 - 71.415) - 0 = -209.415 \, \text{J} \]

Conclusion: Work Done

The total work done by the force on the cow during its displacement is \(-209.415 \, \text{J}\). The negative sign indicates that the force is applied against the direction of the cow's movement.

Key concepts

Integral Calculus in Physics

In physics, integral calculus plays a vital role in analyzing scenarios where forces vary with position. When dealing with variable forces, like the one applied to the cow, we cannot simply use the basic work formula: work is not merely force times distance. Instead, we must integrate the force function over the range of motion.
The work done by a variable force is expressed with the integral \[ W = \int F_x \, dx \] where \( F_x \) is the force component along the direction of motion. This integral accumulates the small contributions of work done as the object moves through each infinitesimal segment of its path.
In the example of the cow, the integration of the variable force from position 0 to 6.9 meters captures the total work done in the process of moving the cow against the opposing force. The result, once calculated, tells us exactly how much energy was expended or gained in moving through the given displacement.

Negative Work

Negative work is an important concept in physics, indicating that the force opposes the motion. It occurs when the direction of the force applied is opposite to the direction of the displacement.
In simpler terms, as seen with the cow, pushing the cow backwards while it is moving away from the barn causes negative work.
Mathematically, negative work occurs in the integration phase, as observed when the net result is negative. This happens because the force function \( F_x = - [20.0 \, \text{N} + (3.0 \, \text{N/m})x] \) results in a negative total due to its opposite direction to the cow's motion.

• The negative sign in work indicates energy is taken out of the system.
• This energy is applied against the object's direction of movement, reducing its total mechanical energy.

Understanding negative work is crucial for identifying systems where forces like friction or tension counteract an object's movement.

Force as a Function of Position

Forces can vary depending on the position of an object, making it necessary to treat these forces as functions of position when calculating work. The force function can describe how the intensity of the force changes as the object moves along a path.
In the case of the exercise, the force applied towards the cow is a function expressed as: \[ F_x = - [20.0 \, \text{N} + (3.0 \, \text{N/m})x] \] This indicates that the force has two components:

• A constant part (-20.0 N), which is uniform across all positions.
• A variable part (-3.0 N/m multiplied by position \( x \)), which increases in strength as the cow moves further away from the starting point.

The force's dependency on position adds complexity to calculating the work done because, rather than being uniform, it requires integration to sum the varying values across the displacement range.
This concept of non-uniform forces is common in physics and must be handled using calculus to ensure accurate results in work calculations.