Problem 15 - University Physics with Modern Physics

Question

On a farm, you are pushing on a stubborn pig with a constant horizontal force with magnitude 30.0 N and direction 37.0\(^\circ\) counterclockwise from the +\(x\)-axis. How much work does this force do during a displacement of the pig that is (a) \(\overrightarrow{s} = (5.00 \, \mathrm{m})\hat{\imath}\); (b) \(\overrightarrow{s} = - (6.00 \, \mathrm{m})\hat{\jmath}\); (c) \(\overrightarrow{s} = - (2.00 \, \mathrm{m})\hat{\jmath} + (4.00 \, \mathrm{m}) \hat{\jmath}\)?

Step-by-Step Solution

Answer

Work results are (a) 120.0 J, (b) -108.0 J, (c) 24.0 J.

Step 1: Calculate Force Components

First, decompose the force into its horizontal and vertical components. The force is 37.0 degrees from the +x-axis.Using trigonometry:Horizontal component (\(F_x\)) = \(30.0 \text{ N} \cdot \cos(37.0^\circ)\)Vertical component (\(F_y\)) = \(30.0 \text{ N} \cdot \sin(37.0^\circ)\)Calculate these values using a calculator:\(F_x = 30.0 \cdot 0.8 = 24.0 \text{ N}\)\(F_y = 30.0 \cdot 0.6 = 18.0 \text{ N}\)

Step 2: Calculate Work for Displacement (a)

For displacement (a), \(\overrightarrow{s} = (5.00 \text{ m}) \hat{\imath}\), we only consider the force component parallel to the displacement.Work is calculated by \(W = F_x \cdot s_x\).Substitute the values:\(W_a = 24.0 \text{ N} \cdot 5.00 \text{ m} = 120.0 \text{ J}\)

Step 3: Calculate Work for Displacement (b)

For displacement (b), \(\overrightarrow{s} = - (6.00 \text{ m}) \hat{\jmath}\), we only consider the vertical component of the force (opposite direction leads to negative work).Work is calculated by \(W = F_y \cdot s_y\).Substitute the values:\(W_b = 18.0 \text{ N} \cdot (-6.00 \text{ m}) = -108.0 \text{ J}\)

Step 4: Calculate Work for Displacement (c)

For displacement (c), \(\overrightarrow{s} = - (2.00 \text{ m}) \hat{\imath} + (4.00 \text{ m}) \hat{\jmath}\), calculate work using both components:\(W_c = F_x \cdot s_x + F_y \cdot s_y\)Substitute values:\(W_c = 24.0 \text{ N} \cdot (-2.00 \text{ m}) + 18.0 \text{ N} \cdot 4.00 \text{ m}\)\(W_c = -48.0 \text{ J} + 72.0 \text{ J} = 24.0 \text{ J}\)

Key Concepts

Vector Components

Understanding vector components is essential in physics, especially when dealing with forces and their impacts in different directions. Any force or displacement in physics can be represented as a vector. A vector has both magnitude and direction, which can be split into perpendicular components — typically along the x and y axes.
Breaking a vector into components helps to analyze how each part of the vector acts along these axes.
To decompose a force like in the original exercise, a common method is using trigonometry:- **Horizontal Component (\(F_x\)):** Multiply the force by the cosine of the angle it forms with the +x-axis.- **Vertical Component (\(F_y\)):** Multiply the force by the sine of the angle.For example, a 30 N force at a 37° angle has a horizontal component of 24 N and a vertical component of 18 N. By breaking down vectors in this way, one can easily determine how much force is acting in any given direction.

Trigonometry in Physics

Trigonometry plays a vital role in physics by providing tools to analyze the angles within different systems. When dealing with vectors, understanding angles helps to resolve forces into useful components.
In physics problems, it's often necessary to determine how a vector aligns with a particular axis. Trigonometry provides sine and cosine functions to determine the projection of a vector onto these axes. For a vector at an angle:- Use \(\cos(\theta)\) to find how much of the vector is aligned with the x-axis.- Use \(\sin(\theta)\) for the y-axis alignment.In our scenario, the vector representing force was 37° from the x-axis. Using trigonometry, the components:

• Horizontal: \(30 \cdot \cos(37°) = 24\text{ N}\)
• Vertical: \(30 \cdot \sin(37°) = 18\text{ N}\)

Knowing these components allows us to calculate how forces impact motion along each axis.

Dot Product

The dot product is a mathematical operation that combines two vectors to yield a scalar. This concept is crucial when calculating work done by a force — specifically when the force is not parallel to the direction of displacement.
The work done by a force can be defined using the dot product equation: \(W = \vec{F} \cdot \vec{s} = |\vec{F}| |\vec{s}| \cos(\theta)\). It incorporates:- Magnitudes of the force and displacement vectors.- The cosine of the angle between the vectors, which accounts for how much of the force actually contributes to work in the direction of displacement.In our problem:

• **Displacement (a):** Work is the horizontal force component multiplied by the horizontal displacement. (No angle needed as they are aligned.)
• **Displacement (b):** Only the vertical component contributes here, giving negative work since they are in opposite directions.
• **Displacement (c):** Involves both components in their respective directions.

The dot product simplifies to the sum of all aligned force and displacement components, making it easier to calculate total work accurately.