/*! 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 ice block floating in a river ... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 8

A ice block floating in a river is pushed through a displacement \(\vec{d}=(20 \mathrm{~m}) \hat{\mathrm{i}}-(16 \mathrm{~m}) \hat{\mathrm{j}}\) along a straight embankment by rushing water, which exerts a force \(\vec{F}=(210 \mathrm{~N}) \hat{\mathrm{i}}-(150 \mathrm{~N}) \hat{\mathrm{j}}\) on the block. How much work does the force do on the block during the displacement?

Short Answer

Expert verified
The work done by the force is 6600 J.

Step by step solution

01

Understanding Work Formula

Work is calculated using the dot product of force and displacement vectors. The formula is given by:\[ W = \vec{F} \cdot \vec{d} = F_x \cdot d_x + F_y \cdot d_y \]where \( W \) is the work performed, \( \vec{F} \) is the force vector, and \( \vec{d} \) is the displacement vector.
02

Identifying Force and Displacement Components

The force vector \( \vec{F} \) is given as \((210\,\text{N}) \hat{\text{i}} - (150\,\text{N}) \hat{\text{j}}\), meaning \( F_x = 210\,\text{N} \) and \( F_y = -150\,\text{N} \). The displacement vector \( \vec{d} \) is given as \((20\,\text{m}) \hat{\text{i}} - (16\,\text{m}) \hat{\text{j}}\), meaning \( d_x = 20\,\text{m} \) and \( d_y = -16\,\text{m} \).
03

Calculating the Dot Product

Use the dot product formula to calculate the work done:\[W = F_x \cdot d_x + F_y \cdot d_y = (210 \times 20) + (-150 \times -16) \]\[ W = 4200 + 2400 \]\[ W = 6600 \, \text{J} \]
04

Result Interpretation

The calculated value indicates that the work done by the force on the block during the displacement is \( 6600 \, \text{J} \) (joules). This means the force has applied energy to move the ice block through the river.

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Key Concepts

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

Force Vector
In physics, a force vector represents a force acting on an object in both magnitude and direction. Vectors are essential in understanding how forces like gravity, friction, or applied forces interact with objects.
When dealing with force vectors, it is crucial to focus on:
  • Magnitude: This tells us how strong the force is. In our exercise, the magnitude of the force vector is expressed through its components in the coordinate axes.
  • Direction: This shows where the force is applied. Vectors have components that align with each axis, helping determine their direction.
For example, if we have a force vector \( \vec{F} = (210 \, \text{N}) \hat{\text{i}} - (150 \, \text{N}) \hat{\text{j}} \), it means the force has a component of 210 N in the positive x-direction and a component of 150 N in the negative y-direction. Understanding force vectors allows us to predict movement caused by different forces acting on a body.
Displacement Vector
Displacement is a vector that measures an object's change in position. Unlike distance, which is scalar, displacement considers both the
  • Starting and ending points
  • Direction of movement
The displacement vector is pivotal in calculating work, as it gives the path along which the force acts.
For instance, in the original exercise, the displacement vector \( \vec{d} = (20\, \text{m}) \hat{\text{i}} - (16\, \text{m}) \hat{\text{j}} \) indicates a shift of 20 meters in the positive x-direction and 16 meters in the negative y-direction.
So when you consider displacement, always look at how both the path and direction influence the outcome of situations involving vectors in physics.
Dot Product
The dot product is a mathematical operation that takes two vectors and returns a scalar, commonly used in physics to compute work. It effectively measures how much one vector (e.g., force) aligns with another (e.g., displacement).
Here's why the dot product is critical:
  • Directional Influence: When vectors are aligned, the dot product maximizes the resultant value. Misalignment reduces this value.
  • Magnitude Calculation: The formula \( W = \vec{F} \cdot \vec{d} = F_x \cdot d_x + F_y \cdot d_y \) sums the product of corresponding components.
In the context of the exercise, calculating the dot product of the given force and displacement vectors, we find:
\[W = (210 \times 20) + (-150 \times -16) \]
yields a total work done of 6600 joules.
Understanding the dot product ensures you can evaluate work in physical systems and appreciate the nuances of vector interactions.

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Most popular questions from this chapter

A spring of spring constant \(5.0 \times 10^{3} \mathrm{~N} / \mathrm{m}\) is stretched initially by \(5.0 \mathrm{~cm}\) from the unstretched position. What is the work required to stretch it further by another \(5.0 \mathrm{~cm}\) ?

Across a horizontal floor, a \(102 \mathrm{~kg}\) block is pulled at a constant speed of \(5.5 \mathrm{~m} / \mathrm{s}\) by an applied force of \(125 \mathrm{~N}\) directed \(38^{\circ}\) above the horizontal. Calculate the rate at which the force does work on the block.

A \(1.0 \mathrm{~kg}\) block is initially at rest on a horizontal frictionless surface when a horizontal force along an \(x\) axis is applied to the block. The force is given by \(\vec{F}(x)=\left(2.5-x^{2}\right) \hat{\mathrm{i}} \mathrm{N}\), where \(x\) is in meters and the initial position of the block is \(x=0\). (a) What is the kinetic energy of the block as it passes through \(x=2.0 \mathrm{~m} ?\) (b) What is the maximum kinetic energy of the block between \(x=0\) and \(x=2.0 \mathrm{~m} ?\)

The only force acting on a \(2.0 \mathrm{~kg}\) canister that is moving in an \(x y\) plane has a magnitude of \(5.0 \mathrm{~N}\). The canister initially has a velocity of \(4.0 \mathrm{~m} / \mathrm{s}\) in the positive \(x\) direction and some time later has a velocity of \(6.0 \mathrm{~m} / \mathrm{s}\) in the positive \(y\) direction. How much work is done on the canister by the \(5.0 \mathrm{~N}\) force during this time?

A \(0.35 \mathrm{~kg}\) ladle sliding on a horizontal frictionless surface is attached to one end of a horizontal spring \((k=450 \mathrm{~N} / \mathrm{m})\) whose other end is fixed. The ladle has a kinetic energy of \(10 \mathrm{~J}\) as it passes through its equilibrium position (the point at which the spring force is zero). (a) At what rate is the spring doing work on the ladle as the ladle passes through its equilibrium position? (b) At what rate is the spring doing work on the ladle when the spring is compressed \(0.10 \mathrm{~m}\) and the ladle is moving away from the equilibrium position?
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