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Chapter 5: Problem 36

\(\mathrm{A}\) force \(\bar{F}=13 \mathrm{~N} \hat{\imath}+13 \mathrm{~N} \hat{\jmath}\) acts on a hockey puck. Determine the work done if the force results in the puck's displacement by \(4.2 \mathrm{~m}\) in the \(+x\) -direction and \(2.1 \mathrm{~m}\) in the \(-y\) -direction.

Short Answer

Expert verified
The work done is 27.3 Joules.

Step by step solution

01

Understand the Concept of Work

The work done by a force is given by the dot product of the force vector and the displacement vector. It is expressed mathematically as \( W = \bar{F} \cdot \bar{d} \), where \( \bar{F} \) is the force vector and \( \bar{d} \) is the displacement vector.
02

Write the Force and Displacement Vectors

The force vector is given as \( \bar{F} = 13 \hat{\imath} + 13 \hat{\jmath} \). The displacement vector, based on the problem description, is \( \bar{d} = 4.2 \hat{\imath} - 2.1 \hat{\jmath} \).
03

Calculate the Dot Product

The dot product \( \bar{F} \cdot \bar{d} \) is calculated component-wise: \[\bar{F} \cdot \bar{d} = (13)(4.2) + (13)(-2.1)\] Simplifying this gives us the work done by the force.
04

Simplify the Expression for Work

Compute each component of the expression from Step 3: - The \( x \)-component: \( 13 \times 4.2 = 54.6 \)- The \( y \)-component: \( 13 \times -2.1 = -27.3 \)Adding these results, the total work done \( W \) is:\[W = 54.6 + (-27.3) = 27.3 \text{ J}\]
05

Conclude with the Result

The work done by the force on the puck, resulting in the given displacement, is 27.3 Joules.

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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 mathematical operation that connects vectors, allowing us to calculate a single scalar value from two vectors. It helps quantify how much of one vector acts in the direction of another.
To perform a dot product, you multiply each of the corresponding components of the vectors and add them up.
  • If you have vectors \( \bar{a} = a_x \hat{\imath} + a_y \hat{\jmath} \) and \( \bar{b} = b_x \hat{\imath} + b_y \hat{\jmath} \), the dot product is given by \( \bar{a} \cdot \bar{b} = a_x b_x + a_y b_y \).
  • The result is a scalar, a single number, not a vector.
This is useful in physics, especially when calculating work, as it checks the alignment of vectors. If vectors are parallel, the dot product is maximized; if perpendicular, it's zero. So, it essentially measures similarity in the direction between two vectors.
Force Vector
In physics, a force vector is a quantity that has both magnitude and direction. It represents a force applied to an object and is expressed in vector notation.
In this problem, the force vector \( \bar{F} \) is given as \( 13 \hat{\imath} + 13 \hat{\jmath} \), which tells us:
  • The magnitude of the force is derived from its components.
  • It exerts 13 N in the positive x-direction and 13 N in the positive y-direction.
Force vectors help determine how forces interact with objects, influencing their motion and energy. Understanding their components is crucial for assessing the actual path or behavior of an object under the force applied.
Displacement Vector
A displacement vector describes the change in position of an object, indicating both the direction and distance.
In this scenario, the puck's displacement vector \( \bar{d} \) is given as \( 4.2 \hat{\imath} - 2.1 \hat{\jmath} \), showing:
  • 4.2 meters in the positive x-direction.
  • -2.1 meters in the negative y-direction.
Displacement vectors are essential in physics for visualizing movement from one point to another. They help integrate the direction of movement, which is key when calculating work using the dot product of force and displacement vectors.
Work Calculation
Calculating work involves using the dot product between the force and displacement vectors. It quantifies the energy transfer when a force causes a displacement.
To find the work done, use the formula:
  • \( W = \bar{F} \cdot \bar{d} \)
In this instance:
  • The force vector \( \bar{F} = 13 \hat{\imath} + 13 \hat{\jmath} \).
  • The displacement vector \( \bar{d} = 4.2 \hat{\imath} - 2.1 \hat{\jmath} \).
  • Multiply corresponding components of these vectors, then sum them: \( 13 \times 4.2 + 13 \times (-2.1) \).
This results in
  • The calculation: \( 54.6 - 27.3 = 27.3 \text{ J} \).
The unit, Joules, measures the energy transferred to or from an object by force, demonstrating how effectively the force contributes to displacement.

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Most popular questions from this chapter

A man normally consumes 8.4 MJ of food energy per day. He then begins running a distance of \(8 \mathrm{~km}\) four times per week. If he expends energy at the rate of \(450 \mathrm{~W}\) while running at \(12 \mathrm{~km} / \mathrm{h},\) how much more food energy should he consume daily in order to maintain constant weight?

A \(24.5-\mathrm{kg}\) boulder that falls from a \(13.4-\mathrm{m}\) cliff strikes the ground with kinetic energy (a) \(3220 \mathrm{~J}\) (b) \(1610 \mathrm{~J}\); (c) \(1450 \mathrm{~J}\) (d) \(328 \mathrm{~J}\).

A Hooke's law spring with \(k=135 \mathrm{~N} / \mathrm{m}\) is compressed \(9.50 \mathrm{~cm}\) from equilibrium. The work required to do this is (a) \(12.8 \mathrm{~J}\) (b) \(1.22 \mathrm{~J} ;\) (c) \(0.61 \mathrm{~J}\); (d) \(0.35 \mathrm{~J}\).

In a bench press, a weight lifter presses a \(105-\mathrm{kg}\) barbell \(0.485 \mathrm{~m}\) straight up. If the barbell starts and ends at rest, how much work did the weight lifter do?

The power required to lift a \(2.85-\mathrm{kg}\) brick \(10.0 \mathrm{~m}\) in \(2.50 \mathrm{~s}\) is (a) \(11.4 \mathrm{~W}\); (b) \(55.9 \mathrm{~W}\) : (c) \(112 \mathrm{~W}\) (d) \(147 \mathrm{~W}\).
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