/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 A loaded grocery cart is rolling... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 8

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

At a waterpark, sleds with riders are sent along a slippery, horizontal surface by the release of a large compressed spring. The spring, with force constant \(k\) = 40.0 N/cm and negligible mass, rests on the frictionless horizontal surface. One end is in contact with a stationary wall. A sled and rider with total mass 70.0 kg are pushed against the other end, compressing the spring 0.375 m. The sled is then released with zero initial velocity. What is the sled's speed when the spring (a) returns to its uncompressed length and (b) is still compressed 0.200 m?

A little red wagon with mass 7.00 kg moves in a straight line on a frictionless horizontal surface. It has an initial speed of 4.00 m/s and then is pushed 3.0 m in the direction of the initial velocity by a force with a magnitude of 10.0 N. (a) Use the work\(-\)energy theorem to calculate the wagon's final speed. (b) Calculate the acceleration produced by the force. Use this acceleration in the kinematic relationships of Chapter 2 to calculate the wagon's final speed. Compare this result to that calculated in part (a).

A 30.0-kg crate is initially moving with a velocity that has magnitude 3.90 m/s in a direction 37.0\(^\circ\) west of north. How much work must be done on the crate to change its velocity to 5.62 m/s in a direction 63.0\(^\circ\) south of east?

On an essentially frictionless, horizontal ice rink, a skater moving at 3.0 m/s encounters a rough patch that reduces her speed to 1.65 m/s due to a friction force that is 25% of her weight. Use the work\(-\)energy theorem to find the length of this rough patch.

To stretch a spring 3.00 cm from its unstretched length, 12.0 J of work must be done. (a) What is the force constant of this spring? (b) What magnitude force is needed to stretch the spring 3.00 cm from its unstretched length? (c) How much work must be done to compress this spring 4.00 cm from its unstretched length, and what force is needed to compress it this distance?
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Nuclear Physics

Read Explanation

Kinematics Physics

Read Explanation

Radiation

Read Explanation

Measurements

Read Explanation

Energy Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.