/*! 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 27 Force on a Particle \(A\) force ... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 27

Force on a Particle \(A\) force \(\vec{F}=(4.0 \mathrm{~N}) \hat{\mathrm{i}}+(c \mathrm{~N}) \hat{\mathrm{j}}\) acts on a particle as the particle goes through displacement \(\vec{d}=(3.0 \mathrm{~m}) \hat{\mathrm{i}}-\) \((2.0 \mathrm{~m}) \hat{\mathrm{j}} .\) (Other forces also act on the particle.) What is the value of \(c\) if the work done on the particle by force \(\vec{F}\) is (a) zero, (b) \(17 \mathrm{~J}\), and (c) \(-18 \mathrm{~J} ?\)

Short Answer

Expert verified
For (a) c = 6.0, for (b) c = -2.5, and for (c) c = 15.

Step by step solution

01

Understand the Problem

We need to find the value of the constant c for three different cases where the work done by the force \(\vec{F}=(4.0 \mathrm{~N}) \hat{\mathrm{i}}+(c \mathrm{~N}) \hat{\mathrm{j}}\) during the displacement \(\vec{d}=(3.0 \mathrm{~m}) \hat{\mathrm{i}}-(2.0 \mathrm{~m}) \hat{\mathrm{j}}\) given the work done is (a) zero, (b) 17 J, and (c) -18 J.
02

Recall the Work Done Formula

The work done by the force \(\vec{F}\) during the displacement \(\vec{d}\) is given by \(\text{Work} = \vec{F} \cdot \vec{d}\). For vectors \(\vec{F} = F_1 \hat{\mathrm{i}} + F_2 \hat{\mathrm{j}} \) and \(\vec{d} = d_1 \hat{\mathrm{i}} + d_2 \hat{\mathrm{j}}\), the dot product is given by \(\vec{F} \cdot \vec{d} = F_1 d_1 + F_2 d_2\).
03

Apply the Given Forces and Displacement

Given \(\vec{F} = (4.0 \mathrm{~N}) \hat{\mathrm{i}} + (c \mathrm{~N}) \hat{\mathrm{j}}\) and \(\vec{d} = (3.0 \mathrm{~m}) \hat{\mathrm{i}} - (2.0 \mathrm{~m}) \hat{\mathrm{j}}\), the dot product becomes \(\vec{F} \cdot \vec{d} = (4.0 \mathrm{~N}) \cdot (3.0 \mathrm{~m}) + (c \mathrm{~N}) \cdot (-2.0 \mathrm{~m})\). Simplifying this, we get \(\vec{F} \cdot \vec{d} = 12.0 - 2c\).
04

Solve for Case (a): Work Done is Zero

Set the work done equal to zero: \(\12.0 - 2c = 0\). Solving for c, we find \(\12.0 = 2c\), hence \(\c = 6.0\).
05

Solve for Case (b): Work Done is 17 J

Set the work done equal to 17 J: \(\12.0 - 2c = 17\). Solving for c, we get \(\12.0 - 17 = 2c\) which simplifies to \(\-5 = 2c\), hence \(\c = -2.5\).
06

Solve for Case (c): Work Done is -18 J

Set the work done equal to -18 J: \(\12.0 - 2c = -18\). Solving for c, we have \(\12.0 + 18 = 2c\) which simplifies to \(\30 = 2c\), hence \(\c = 15\).

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Key Concepts

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

Dot Product
The dot product is a way to multiply two vectors, resulting in a scalar. It helps us find out how much one vector *points* in the same direction as another. To find the dot product, we use the formula \( \vec{A} \cdot \vec{B} = A_1 B_1 + A_2 B_2 \), where \vec{A} = A_1 \hat{i} + A_2 \hat{j}\ and \vec{B} = B_1 \hat{i} + B_2 \hat{j}\. This means we multiply the corresponding components of the vectors and then sum them up.
Force and Displacement
When we talk about work done by a force, we're usually thinking about how a force causes an object to move. The relationship between force and displacement is crucial because the work done is the product of these two factors. Consider a force \(\vec{F} = 4.0 \mathrm{N} \hat{i} + c \mathrm{N} \hat{j}\) and a displacement \(\vec{d} = 3.0 \mathrm{m} \hat{i} - 2.0 \mathrm{m} \hat{j}\). The components of force and displacement interact through the dot product to result in the work done. This shows the importance of understanding both vectors鈥� directions and magnitudes.
Vector Mathematics
Vectors are quantities that have both magnitude and direction, and vector mathematics involves many operations such as addition, subtraction, and multiplication by a scalar or dot product. For example, displacement is a vector (where the object moves), and so is force (which causes the motion). Working with vectors requires precise calculations to understand physical phenomena correctly. For instance, to solve for work done using vectors, we first find the dot product, which simplifies the relationship between force and displacement into a scalar quantity.
Physics Problems
Physics problems often involve calculating how different forces influence objects. In the exercise, calculating the work done by a force on a particle as it moves through a given displacement incorporates multiple physics concepts including vector mathematics, dot products, and the physical significance of work. For example, if the work done is zero, the force is perpendicular to the displacement. When the work done is positive or negative, it shows the force doing work on the particle to move it in its direction or against it, respectively. Thus, understanding these concepts helps in solving such real-world physics problems efficiently.

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Most popular questions from this chapter

Object Accelerates Horizontally An initially stationary \(2.0 \mathrm{~kg}\) object accelerates horizontally and uniformly to a speed of \(10 \mathrm{~m} / \mathrm{s}\) in \(3.0 \mathrm{~s}\). (a) In that \(3.0 \mathrm{~s}\) interval, how much work is done on the object by the force accelerating it? What is the instantaneous power due to that force (b) at the end of the interval and (c) at the end of the first half of the interval?

A Particle Moves A particle moves along a straight path through displacement \(\vec{d}=(8 \mathrm{~m}) \hat{\mathrm{i}}-(c \mathrm{~m}) \hat{\mathrm{j}}\) while force \(\vec{F}=\) \((2 \mathrm{~N}) \mathrm{i}-(4 \mathrm{~N}) \hat{\mathrm{j}}\) acts on it. (Other forces also act on the particle.) What is the value of \(c\) if the work done by \(\vec{F}\) on the particle is (a) zero, (b) positive, and (c) negative?

Cold Hot Dogs Figure \(9-32\) shows a cold package of hot dogs sliding rightward across a frictionless floor through a distance \(d=\) \(20.0 \mathrm{~cm}\) while three forces are applied to it. Two of the forces are horizontal and have the magnitudes \(F_{A}=5.00 \mathrm{~N}\) and \(F_{B}=1.00 \mathrm{~N} ;\) the third force is angled down by \(\theta=-60.0^{\circ}\) and has the magnitude \(F_{C}=4.00 \mathrm{~N}\). (a) For the \(20.0 \mathrm{~cm}\) displacement, what is the net work done on the package by the three applied forces, the gravitational force on the package, and the normal force on the package? (b) If the package has a mass of \(2.0 \mathrm{~kg}\) and an initial kinetic energy of \(0 \mathrm{~J}\), what is its speed at the end of the displacement?

Electron in Copper If an electron (mass \(m=9.11 \times 10^{-31} \mathrm{~kg}\) ) in copper near the lowest possible temperature has a kinetic energy of \(6.7 \times 10^{-19} \mathrm{~J}\), what is the speed of the electron?

Work Done by \(2-\mathrm{D}\) Force What work is done by a force \(\vec{F}=\) \(((2 \mathrm{~N} / \mathrm{m}) x) \hat{\mathrm{i}}+(3 \mathrm{~N}) \hat{j}\), with \(x\) in meters, that moves a particle from a position \(\vec{r}_{1}=(2 \mathrm{~m}) \hat{\mathrm{i}}+(3 \mathrm{~m}) \hat{\mathrm{j}}\) to a position \(\vec{r}_{2}=-(4 \mathrm{~m}) \hat{\mathrm{i}}-(3 \mathrm{~m}) \hat{\mathrm{j}}\) ?
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