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A force \(\vec{F}=(4.0 \mathrm{~N}) \hat{\mathrm{i}}+c \hat{\mathrm{j}}\) 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 \(c\) if the work done on the particle by force \(\vec{F}\) is \((\mathrm{a}) 0,(\mathrm{~b}) 17 \mathrm{~J},\) and \((\mathrm{c})-18 \mathrm{~J} ?\)

Short Answer

Expert verified
(a) \(c = 6\); (b) \(c = -2.5\); (c) \(c = 15\).

Step by step solution

01

Understanding the Problem

The problem asks us to find the constant \( c \) in the force vector \( \vec{F} = (4.0 \mathrm{~N}) \hat{\mathrm{i}} + c \hat{\mathrm{j}} \) such that the work done by this force on the displacement \( \vec{d} = (3.0 \mathrm{~m}) \hat{\mathrm{i}} - (2.0 \mathrm{~m}) \hat{\mathrm{j}} \) is a specific value (0, 17 J, and -18 J).
02

Formula for Work Done by a Force

The work done by a force \( \vec{F} \) on a displacement \( \vec{d} \) is given by the dot product: \( W = \vec{F} \cdot \vec{d} \). The formula for the dot product is \( \vec{F} \cdot \vec{d} = F_x \cdot d_x + F_y \cdot d_y \) where \( F_x \) and \( F_y \) are the components of the force, and \( d_x \) and \( d_y \) are the components of the displacement.
03

Expressing the Dot Product

Substituting the given vectors into the dot product formula, we have: \( W = (4.0) \cdot (3.0) + c \cdot (-2.0) = 12 - 2c \). This expression represents the work done based on the parameter \( c \).
04

Solving for Condition (a) Work Done is 0 J

For \( W = 0 \), substitute into the equation: \( 12 - 2c = 0 \). Solving for \( c \), \( 2c = 12 \) leads to \( c = 6 \).
05

Solving for Condition (b) Work Done is 17 J

For \( W = 17 \), substitute into the equation: \( 12 - 2c = 17 \). Solving for \( c \), \( 2c = 12 - 17 \) gives \( 2c = -5 \). Thus, \( c = -2.5 \).
06

Solving for Condition (c) Work Done is -18 J

For \( W = -18 \), substitute into the equation: \( 12 - 2c = -18 \). Solving for \( c \), \( 2c = 12 + 18 \) results in \( 2c = 30 \). Therefore, \( c = 15 \).

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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, also known as the scalar product, is an essential concept in physics and mathematics. It's a way to multiply two vectors, resulting in a scalar quantity. This scalar shows the extent to which one vector extends in the direction of another.

For two vectors, \( \vec{A} = a_x\hat{i} + a_y\hat{j} \) and \( \vec{B} = b_x\hat{i} + b_y\hat{j} \), the dot product is calculated as:
  • \( \vec{A} \cdot \vec{B} = a_x b_x + a_y b_y \)
This operation is widely used to determine the work done by a force when it's applied over a displacement. In such cases, it reflects the component of force that acts in the direction of the displacement, which is crucial for calculating work performed.

In the given problem, understanding and using the dot product enable us to find the unknown parameter \( c \) in the force vector amid various work conditions.
Force Vector
A force vector in physics is a representation of force with both magnitude and direction. Typically expressed in terms of unit vectors like \( \hat{i} \) and \( \hat{j} \), it helps in breaking down the force into its horizontal and vertical components.

In the exercise example, the force vector \( \vec{F} = (4.0 \mathrm{~N}) \hat{\mathrm{i}} + c \hat{\mathrm{j}} \) indicates:
  • 4.0 N of force is being applied in the horizontal direction (\( \hat{i} \)).
  • A variable \( c \) describes the force component in the vertical direction (\( \hat{j} \)).
By analyzing this vector, students can understand how forces are applied and how they influence movement, crucial for solving various physics problems involving vectors.
Displacement Vector
The displacement vector connects the initial and final position of a moving particle, showing the direction and distance of movement. Like forces, displacements can be broken down into components using unit vectors.

In our problem, the displacement is characterized by \( \vec{d} = (3.0 \mathrm{~m}) \hat{\mathrm{i}} - (2.0 \mathrm{~m}) \hat{\mathrm{j}} \):
  • The horizontal component (3.0 m \( \hat{i} \)) is the distance moved right.
  • The vertical component (-2.0 m \( \hat{j} \)) shows downward movement.
Grasping the concept of displacement vectors is essential because they play a critical role when computing the work done by forces acting upon objects, providing insight into the vectorial nature of movement.
Solving Physics Problems
When tackling physics problems, a structured approach often leads to success. An example is the solution to this exercise, which asks to find values of \( c \) for the force vector given various work values.

Follow these steps to solve related problems effectively:
  • Understand the Problem: Break down the given information, like force and displacement vectors.
  • Apply Relevant Formulas: Utilize the dot product to link force and displacement components to calculate work.
  • Solve for Variables: Substitute known values into the equations to find unknown variables like \( c \).
  • Verify Solutions: Check calculations to ensure the results align with physics principles.
This systematic approach, emphasizing understanding each component such as force and displacement vectors and their interaction through the dot product, leads to efficient and correct problem-solving.

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Most popular questions from this chapter

An explosion at ground level leaves a crater with a diameter that is proportional to the energy of the explosion raised to the \(\frac{1}{3}\) power; an explosion of 1 megaton of TNT leaves a crater with a \(1 \mathrm{~km}\) diameter. Below Lake Huron in Michigan there appears to be an ancient impact crater with a \(50 \mathrm{~km}\) diameter. What was the kinetic energy associated with that impact, in terms of (a) megatons of TNT (1 megaton yields \(4.2 \times 10^{15} \mathrm{~J}\) ) and (b) Hiroshima bomb equivalents (13 kilotons of TNT each)? (Ancient meteorite or comet impacts may have significantly altered the climate, killing off the dinosaurs and other life-forms.)

A \(45 \mathrm{~kg}\) block of ice slides down a frictionless incline \(1.5 \mathrm{~m}\) long and \(0.91 \mathrm{~m}\) high. A worker pushes up against the ice, parallel to the incline, so that the block slides down at constant speed. (a) Find the magnitude of the worker's force. How much work is done on the block by (b) the worker's force, (c) the gravitational force on the block, (d) the normal force on the block from the surface of the incline, and (e) the net force on the block?

A can of sardines is made to move along an \(x\) axis from \(x=0.25 \mathrm{~m}\) to \(x=1.25 \mathrm{~m}\) by a force with a magnitude given by \(F=\exp \left(-4 x^{2}\right),\) with \(x\) in meters and \(F\) in newtons. (Here exp is the exponential function.) How much work is done on the can by the force?

A proton (mass \(m=1.67 \times 10^{-27} \mathrm{~kg}\) ) is being accelerated along a straight line at \(3.6 \times 10^{15} \mathrm{~m} / \mathrm{s}^{2}\) in a machine. If the proton has an initial speed of \(2.4 \times 10^{7} \mathrm{~m} / \mathrm{s}\) and travels \(3.5 \mathrm{~cm},\) what then is (a) its speed and (b) the increase in its kinetic energy?

(a) At a certain instant, a particle-like object is acted on by a force \(\vec{F}=(4.0 \mathrm{~N}) \hat{\mathrm{i}}-(2.0 \mathrm{~N}) \hat{\mathrm{j}}+(9.0 \mathrm{~N}) \hat{\mathrm{k}}\) while the object's velocity is \(\vec{v}=-(2.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(4.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{k}} .\) What is the instantaneous rate at which the force does work on the object? (b) At some other time, the velocity consists of only a \(y\) component. If the force is unchanged and the instantaneous power is \(-12 \mathrm{~W},\) what is the velocity of the object?

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