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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

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

Step by step solution

01

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 \( W = \overrightarrow{F} \cdot \overrightarrow{s} \).
02

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 \( \overrightarrow{A} \cdot \overrightarrow{B} = a_1b_1 + a_2b_2 \).
03

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 \( W = 30(-9.0) + (-40)(3.0) \).
04

Calculate the Components

Calculate each component separately: - For the \( \hat{\imath} \) component: \( 30 \times -9.0 = -270 \)- For the \( \hat{\jmath} \) component: \((-40) \times 3.0 = -120 \).
05

Sum the Components

Add the two results obtained from the components: - \( -270 + (-120) = -390 \).
06

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

You are asked to design spring bumpers for the walls of a parking garage. A freely rolling 1200-kg car moving at 0.65 m/s is to compress the spring no more than 0.090 m before stopping. What should be the force constant of the spring? Assume that the spring has negligible mass.

An airplane in flight is subject to an air resistance force proportional to the square of its speed v. But there is an additional resistive force because the airplane has wings. Air flowing over the wings is pushed down and slightly forward, so from Newton's third law the air exerts a force on the wings and airplane that is up and slightly backward (\(\textbf{Fig. P6.94}\)). The upward force is the lift force that keeps the airplane aloft, and the backward force is called \(induced \, drag\). At flying speeds, induced drag is inversely proportional to \(v^2\), so the total air resistance force can be expressed by \(F_air = \alpha v^{2} + \beta /v{^2}\), where \(\alpha\) and \(\beta\) are positive constants that depend on the shape and size of the airplane and the density of the air. For a Cessna 150, a small single-engine airplane, \(\alpha = 0.30 \, \mathrm{N} \cdot \mathrm{s^{2}/m^{2}}\) and \(\beta = 3.5 \times 10^5 \, \mathrm{N} \cdot \mathrm{m^2/s^2}\). In steady flight, the engine must provide a forward force that exactly balances the air resistance force. (a) Calculate the speed (in km/h) at which this airplane will have the maximum \(range\) (that is, travel the greatest distance) for a given quantity of fuel. (b) Calculate the speed (in km/h) for which the airplane will have the maximum \(endurance\)(that is, remain in the air the longest time).

You push your physics book 1.50 m along a horizontal tabletop with a horizontal push of 2.40 N while the opposing force of friction is 0.600 N. How much work does each of the following forces do on the book: (a) your 2.40-N push, (b) the friction force, (c) the normal force from the tabletop, and (d) gravity? (e) What is the net work done on the book?

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?

Using a cable with a tension of 1350 N, a tow truck pulls a car 5.00 km along a horizontal roadway. (a) How much work does the cable do on the car if it pulls horizontally? If it pulls at 35.0\(^\circ\) above the horizontal? (b) How much work does the cable do on the tow truck in both cases of part (a)? (c) How much work does gravity do on the car in part (a)?

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