Problem 8 - University Physics with Modern Physics

Question

A loaded grocery cart is rolling across a parking lot in a strong wind. You apply a constant force \(\vec{F}=(30 \mathrm{N}) \hat{\imath}-(40 \mathrm{N}) \hat{\mathrm{J}}\) to the cart as it undergoes a displacement \(\vec{s}=(-9.0 \mathrm{m}) \hat{\boldsymbol{\imath}}-(3.0 \mathrm{m}) \hat{\boldsymbol{J}}\) . How much work does the force you apply do on the grocery cart?

Step-by-Step Solution

Answer

The work done on the grocery cart is -150 Joules.

Step 1: Understand the Problem

We are tasked to find the work done by a force applied to a grocery cart that moves with a given displacement. The force and displacement are both given as vectors in component form.

Step 2: Recall the Work Formula

The work done by a force \((\vec{F})\) when an object undergoes a displacement \((\vec{s})\) is given by the formula: \[ W = \vec{F} \cdot \vec{s} \] where ' \cdot \ ' denotes the dot product of two vectors.

Step 3: Write Down the Force and Displacement Vectors

Given \( \vec{F} = (30 \mathrm{~N}) \hat{\imath} - (40 \mathrm{~N}) \hat{\mathrm{J}} \) and \( \vec{s} = (-9.0 \mathrm{~m}) \hat{\imath} - (3.0 \mathrm{~m}) \hat{\boldsymbol{J}} \). These represent the components of force and displacement in the i (horizontal) and j (vertical) directions, respectively.

Step 4: Compute the Dot Product

The dot product is computed as follows: \[ \vec{F} \cdot \vec{s} = (F_x \cdot s_x) + (F_y \cdot s_y) \] Substitute the given values: \[ (30 \cdot -9) + (-40 \cdot -3) = -270 + 120 \]

Step 5: Calculate the Work Done

Now, calculate the actual work value: \[ W = -270 + 120 = -150 \text{ J} \] Thus, the work done by the force on the grocery cart is -150 Joules.

Key Concepts

Understanding the Dot Product

In physics and mathematics, the dot product is a method to multiply two vectors, yielding a scalar quantity. This operation is an essential mathematical tool when dealing with vectors, particularly in calculating work done by a force. To compute the dot product of two vectors, one should perform the following steps:

• Multiply the corresponding components of the vectors together.
• Add the results of these multiplications.

For instance, given the force vector \( \vec{F}=(30\mathrm{N}) \hat{\imath}-(40\mathrm{N}) \hat{\mathrm{J}} \) and the displacement vector \( \vec{s}=(-9.0\mathrm{m}) \hat{\imath}-(3.0\mathrm{m}) \hat{\boldsymbol{J}} \), the dot product is calculated as:
\[ \vec{F} \cdot \vec{s} = (30 \cdot -9) + (-40 \cdot -3) = -270 + 120 \]
The calculated dot product is \(-150\). This value is crucial as it tells us about the work done when the force and displacement are considered in relation to each other.The sign of the dot product (positive or negative) indicates the nature of work done concerning direction.
A positive value means the force supports the displacement, while a negative value (as in our example) suggests the force opposes the displacement.

Force and Displacement in Physics Problems

In physics, force is any interaction that, when unopposed, changes the motion of an object. Displacement, on the other hand, refers to the change in position of an object. When looking to calculate work done in physics, force and displacement are intertwined concepts. Both are represented as vectors, meaning they have both magnitude and direction.

The importance of direction becomes clear when computing work. The formula to find work \(W\) is:
\[ W = \vec{F} \cdot \vec{s} \]
This implies that only the component of the force that acts in the direction of displacement contributes to the work. In the given problem, you applied a constant force to a grocery cart with vectors \( \vec{F}=(30~\mathrm{N}) \hat{\imath}-(40~\mathrm{N}) \hat{\mathrm{J}} \) and displacement vector \( \vec{s}=(-9.0~\mathrm{m}) \hat{\imath}-(3.0~\mathrm{m}) \hat{\boldsymbol{J}} \).

The solution to the work problem requires identifying both the magnitude and directions of these vectors. When tackling real-world physics problems, always pay attention not just to the size of the vectors but also to their orientation in space, as this significantly impacts work done.

Breaking Down Vector Components

Vectors like force and displacement in physics are often broken down into their components for ease of understanding and computation. This decomposition is typically into x, y, and sometimes z components, corresponding to horizontal, vertical, and another direction if needed.

In this specific exercise, both the force and displacement vectors were given in terms of their component forms along the i and j directions:

• \( \vec{F} = (30~\mathrm{N}) \hat{\imath} - (40~\mathrm{N}) \hat{\mathrm{J}} \)
• \( \vec{s} = (-9.0~\mathrm{m}) \hat{\imath} - (3.0~\mathrm{m}) \hat{\boldsymbol{J}} \)

Each component represents the effect of the vector in that particular direction. For example:

• The i-component (\( \hat{\imath} \)) represents the horizontal effect or direction.
• The j-component (\( \hat{\boldsymbol{J}} \)) represents the vertical effect or direction.

When vectors are broken down into their components, it becomes straightforward to perform calculations such as finding the dot product, which is used to determine work.
Understanding vector components helps in visualizing and solving problems efficiently, ensuring a comprehensive approach to vector-related tasks in physics.