Question

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

Short answer

The work done by the force on the cart is \(-390 \mathrm{J}\).

Step-by-step solution

Identify the Formula

The work done by a constant force is given by the dot product of the force vector \( \overrightarrow{F} \) and the displacement vector \( \overrightarrow{s} \). Mathematically, this is expressed as \( W = \overrightarrow{F} \cdot \overrightarrow{s} \).

Apply the Dot Product Formula

The dot product of two vectors \( \overrightarrow{A} = a_1 \hat{\imath} + a_2 \hat{\jmath} \) and \( \overrightarrow{B} = b_1 \hat{\imath} + b_2 \hat{\jmath} \) is calculated as \( \overrightarrow{A} \cdot \overrightarrow{B} = a_1b_1 + a_2b_2 \).

Substitute the Values

Substitute the given values of the force vector \( \overrightarrow{F} = (30 \, \mathrm{N})\hat{\imath} - (40 \, \mathrm{N})\hat{\jmath} \) and the displacement vector \( \overrightarrow{s} = (-9.0 \, \mathrm{m})\hat{\imath} + (3.0 \, \mathrm{m})\hat{\jmath} \) into the dot product formula. This means calculating \( W = 30(-9.0) + (-40)(3.0) \).

Calculate the Components

Calculate each component separately: - For the \( \hat{\imath} \) component: \( 30 \times -9.0 = -270 \)- For the \( \hat{\jmath} \) component: \((-40) \times 3.0 = -120 \).

Sum the Components

Add the two results obtained from the components: - \( -270 + (-120) = -390 \).

State the Result

The total work done by the force is \(-390 \mathrm{J}\). The negative sign indicates that the force applied is opposite to the direction of displacement.

Key concepts

Understanding the Dot Product

The dot product is a mathematical tool used when dealing with vectors. It's particularly useful when we need to calculate work, as in the problem of the rolling grocery cart.

• What is a dot product? The dot product of two vectors results in a scalar, which is a single number, not a vector.
• How is it used? When you have two vectors expressed in a Cartesian coordinate system, such as \(\overrightarrow{F} = (a_1)\hat{\imath} + (a_2)\hat{\jmath}\) and \(\overrightarrow{s} = (b_1)\hat{\imath} + (b_2)\hat{\jmath}\), the dot product is calculated as \(a_1b_1 + a_2b_2\).
• Why use it for work calculations? In physics, work is defined as the force applied in the direction of the displacement. The dot product naturally incorporates this directional component, providing the magnitude of work done by the force.

The grocery cart exercise showcases how the dot product simplifies calculations involving force and displacement. You multiply the corresponding components of two vectors and sum them up, yielding energy used or needed.

Force and Displacement in Work Calculations

Force and displacement are the heart of work and energy calculations. Understanding how they interact is vital when solving physics problems.

• Force: This is any interaction that, when unopposed, changes the motion of an object. It's a vector quantity, meaning it has both a magnitude and a direction. In the exercise, the force vector is given as \(\overrightarrow{F} = (30 \, \mathrm{N})\hat{\imath} - (40 \, \mathrm{N})\hat{\jmath}\).
• Displacement: This is the change in position of an object. Like force, it's also a vector. For the rolling cart, the displacement is \(\overrightarrow{s} = (-9.0 \, \mathrm{m})\hat{\imath} + (3.0 \, \mathrm{m})\hat{\jmath}\).
• Relationship in Work: Work occurs when a force causes displacement. The amount of work done is determined by both the magnitude of the force and how much it contributes to moving an object along the direction of displacement.

Through understanding these vectors and their components, you'll realize how vital each part is in determining the net work done, as shown in the problem's step-by-step solution.

Mastering Vector Mathematics

Vector mathematics is essential for solving many physics problems, especially those involving forces and motions, like the grocery cart problem.

• Vector Basics: A vector represents a quantity with both a magnitude and a direction. In mathematical terms, they are often expressed in components, using unit vectors such as \(\hat{\imath}\) and \(\hat{\jmath}\).
• Components and Notation: Each vector component corresponds to a dimension (e.g., horizontal and vertical). For example, \(\overrightarrow{F} = (30 \, \mathrm{N})\hat{\imath} - (40 \, \mathrm{N})\hat{\jmath}\) shows force components along the x and y axes.
• Mathematical Operations: Vectors can be added, subtracted, and multiplied. Multiplication includes the dot product, which is key in calculating work, as it takes into account the directional agreement between force and displacement.

Understanding vector operations and notation helps simplify complex physics problems. By practicing these operations, you'll find that tasks such as calculating work from vectors become much more manageable.