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A constant force \(\overrightarrow{\mathbf{F}}=(2.0 \hat{\mathbf{i}}+4.0 \hat{\mathbf{j}}) \mathbf{N}\) acts on an object as it moves along a straight- line path. If the object's displacement is \(\overrightarrow{\mathbf{d}}=(1.0 \hat{\mathbf{i}}+5.0 \mathbf{j}) \mathrm{m},\) calculate the work done by \(\overrightarrow{\mathbf{F}}\) using these alternate ways of writing the dot product: (a) \(W=F d \cos \theta\) \(;\) (b) \(W=F_{x} d_{x}+F_{y} d_{y}\)

Short Answer

Expert verified
The work done by the force is 22 J.

Step by step solution

01

Understanding the formula for work

Work is calculated by the formula: \[ W = \overrightarrow{\mathbf{F}} \cdot \overrightarrow{\mathbf{d}} = Fd \cos \theta \]where \(W\) is the work done, \(F\) is the magnitude of the force, \(d\) is the magnitude of the displacement, and \(\theta\) is the angle between the force and displacement vectors.
02

Calculate the magnitudes of Force and Displacement

First, calculate the magnitudes of the force \(\overrightarrow{\mathbf{F}} = (2.0 \hat{\mathbf{i}} + 4.0 \hat{\mathbf{j}}) \mathbf{N}\) and the displacement \(\overrightarrow{\mathbf{d}} = (1.0 \hat{\mathbf{i}} + 5.0 \mathbf{j}) \mathrm{m}\).Magnitude of force \(F\) is:\[ F = \sqrt{(2.0)^2 + (4.0)^2} = \sqrt{4 + 16} = \sqrt{20} \approx 4.47 \text{ N} \]Magnitude of displacement \(d\) is:\[ d = \sqrt{(1.0)^2 + (5.0)^2} = \sqrt{1 + 25} = \sqrt{26} \approx 5.10 \text{ m} \]
03

Calculate the angle \(\theta\)

Use the dot product formula to find \(\cos \theta\):\[ \overrightarrow{\mathbf{F}} \cdot \overrightarrow{\mathbf{d}} = Fd \cos \theta \]Calculate the dot product separately:\[ \overrightarrow{\mathbf{F}} \cdot \overrightarrow{\mathbf{d}} = (2.0)(1.0) + (4.0)(5.0) = 2 + 20 = 22 \]Thus,\[ 22 = 4.47 \times 5.10 \times \cos \theta \]Therefore, solve for \(\cos \theta\):\[ \cos \theta = \frac{22}{4.47 \times 5.10} \approx 0.97 \]
04

Verify using formula (b)

Calculate the work done using component-wise dot product \[ W = F_{x} d_{x} + F_{y} d_{y} \]where \(F_x = 2.0\), \(F_y = 4.0\), \(d_x = 1.0\), \(d_y = 5.0\).Substitute the values:\[ W = (2.0)(1.0) + (4.0)(5.0) = 2 + 20 = 22 \text{ J}\]
05

Conclusion

Both methods (a) and (b) provide consistent results showing that the work done by the force \(\overrightarrow{\mathbf{F}}\) on the object during the displacement \(\overrightarrow{\mathbf{d}}\) is 22 J.

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Key Concepts

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

Dot Product
In physics, understanding the dot product is crucial for computing work done by a force. Intuitively, the dot product gives a way of combining two vectors, which reflects how much one vector goes in the direction of another. This concept becomes particularly relevant when calculating work, as it encapsulates the interplay between force and motion.
The dot product of vectors \( \overrightarrow{\mathbf{F}} = (F_x \hat{\mathbf{i}} + F_y \hat{\mathbf{j}}) \) and \( \overrightarrow{\mathbf{d}} = (d_x \hat{\mathbf{i}} + d_y \hat{\mathbf{j}}) \) is given by:
  • \( \overrightarrow{\mathbf{F}} \cdot \overrightarrow{\mathbf{d}} = F_x d_x + F_y d_y \)
This formula shows how the force components pair with their corresponding displacement components.
Alternatively, the dot product can be expressed in terms of the magnitudes of the vectors and the cosine of the angle \( \theta \) between them:
  • \( W = F d \cos \theta \)
Here, \( F \) and \( d \) are the magnitudes of the force and displacement respectively. By accounting for the directionality of force, the dot product hence provides an effective tool to measure work.
Force and Displacement
Work requires both force and displacement. In this problem, we have a force vector and a displacement vector given in terms of their components. The force \( \overrightarrow{\mathbf{F}} = 2.0 \hat{\mathbf{i}} + 4.0 \hat{\mathbf{j}} \) N acts on an object that moves along a displacement \( \overrightarrow{\mathbf{d}} = 1.0 \hat{\mathbf{i}} + 5.0 \hat{\mathbf{j}} \) m.
Understanding these vectors means breaking them down in terms of their horizontal (i) and vertical (j) components. The work done by the force is calculated by evaluating the dot product of these vectors. It represents the energy transferred from the force to the object as it travels from one position to another. The closer the force is aligned with the direction of displacement, the more work it does. If the force were perpendicular to the displacement, the work done would be zero, as the dot product would be zero.
In simpler terms:
  • Force and displacement must "point" in similar directions for work to be done.
  • The magnitude of their alignment defines the work value.
Vector Components
Vectors in physics describe quantities with magnitude and direction. Force and displacement, both being vector quantities, can be described in terms of their components which lie along designated axes, usually represented as the i (horizontal) and j (vertical) axes.
For example, the force vector \( \overrightarrow{\mathbf{F}} = 2.0 \hat{\mathbf{i}} + 4.0 \hat{\mathbf{j}} \) N is split between its i and j components:
  • i-component: 2.0 N
  • j-component: 4.0 N
Similarly, the displacement vector \( \overrightarrow{\mathbf{d}} = 1.0 \hat{\mathbf{i}} + 5.0 \hat{\mathbf{j}} \) m is made of:
  • i-component: 1.0 m
  • j-component: 5.0 m
The key to using vector components effectively is understanding that the dot product formula essentially multiplies each component of force with the corresponding component of displacement. By doing this, we find how much of the force effectively 'does work' in the direction of each portion of the displacement, bridging the gap between the conceptual understanding and mathematical calculation of physics phenomena.

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Most popular questions from this chapter

Stretchable ropes are used to safely arrest the fall of rock climbers. Suppose one end of a rope with unstretched length \(\ell\) is anchored to a cliff and a climber of mass \(m\) is attached to the other end. When the climber is a height \(\ell\) above the anchor point, he slips and falls under the influence of gravity for a distance 2\(\ell\) , after which the rope becomes taut and stretches a distance \(x\) as it stops the climber (see Fig. \(33 ) .\) Assume a stretchy rope behaves as a spring with spring constant \(k . ( a )\) Applying the work-energy principle, show that \(x = \frac { m g } { k } \left[ 1 + \sqrt { 1 + \frac { 4 k \ell } { m g } } \right]\) (b) Assuming \(m = 85 \mathrm { kg } , \quad \ell = 8.0 \mathrm { m }\) and \(k = 850 \mathrm { N } / \mathrm { m } ,\) determine \(x / \ell\) (the fractional stretch of the rope) and \(k x / m g\) (the force that the rope exerts on the climber compared to his own weight) at climber's fall has been stopped.

A 3.0 -m-long steel chain is stretched out along the top level of a horizontal scaffold at a construction site, in such a way that \(2.0 \mathrm{~m}\) of the chain remains on the top level and \(1.0 \mathrm{~m}\) hangs vertically, Fig. \(7-26 .\) At this point, the force on the hanging segment is sufficient to pull the entire chain over the edge. Once the chain is moving, the kinetic friction is so small that it can be neglected. How much work is performed on the chain by the force of gravity as the chain falls from the point where \(2.0 \mathrm{~m}\) remains on the scaffold to the point where the entire chain has left the scaffold? (Assume that the chain has a linear weight density of \(18 \mathrm{~N} / \mathrm{m}\).)

Two forces, \(\overrightarrow{\mathbf{F}}_{1}=(1.50 \hat{\mathbf{i}}-0.80 \hat{\mathbf{j}}+0.70 \hat{\mathbf{k}}) \mathbf{N}\) and \(\overrightarrow{\mathbf{F}}_{2}=\) \((-0.70 \hat{i}+1.20 \hat{j}) \mathbf{N},\) are applied on a moving object of mass \(0.20 \mathrm{~kg} .\) The displacement vector produced by the two forces is \(\mathbf{d}=(8.0 \hat{\mathbf{i}}+6.0 \mathbf{j}+5.0 \hat{\mathbf{k}}) \mathrm{m} .\) What is the work done by the two forces?

A 46.0-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of \(225 \mathrm{~N}\). For the first \(11.0 \mathrm{~m}\) the floor is frictionless, and for the next \(10.0 \mathrm{~m}\) the coefficient of friction is \(0.20 .\) What is the final speed of the crate after being pulled these \(21.0 \mathrm{~m} ?\)

A 3.0 -g locust reaches a speed of \(3.0 \mathrm{~m} / \mathrm{s}\) during its jump. What is its kinetic energy at this speed? (b) If the locust transforms energy with \(35 \%\) efficiency, how much energy is required for the jump?

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