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Chapter 7: Problem 1

A fisherman reels in \(12.0 \mathrm{~m}\) of line while landing a fish, using a constant forward pull of \(25.0 \mathrm{~N}\). How much work does the tension in the line do on the fish?

Short Answer

Expert verified
The work done is 300 J.

Step by step solution

01

Understanding the Work Formula

Work is defined as the product of force and the displacement in the direction of the force. The formula to calculate work done is \( W = F \cdot d \cdot \cos(\theta) \), where \( W \) is work, \( F \) is force, \( d \) is displacement, and \( \theta \) is the angle between the force and displacement vectors.
02

Evaluate the Given Data

The fisherman uses a constant force of \( 25.0 \, \text{N} \) and reels in \( 12.0 \, \text{m} \) of line. The force and the displacement are in the same direction, meaning \( \theta = 0 \). The cosine of 0 degrees is 1, which simplifies our formula.
03

Substitute and Simplify

Substituting the known values into the work formula, we have: \( W = 25.0 \, \text{N} \times 12.0 \, \text{m} \times \cos(0) \). Since \( \cos(0) = 1 \), this simplifies to \( W = 25.0 \, \text{N} \times 12.0 \, \text{m} \).

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Key Concepts

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

Work and Energy
When tackling physics problems related to work and energy, it's essential to grasp the basic definition of work. Work occurs when a force causes a displacement. In simpler terms, it's all about making objects move by applying a force.
To mathematically express work, we use the formula:
  • Work (\( W \)) is the product of the applied force (\( F \)) and the displacement (\( d \)) in the force's direction, multiplied by the cosine of the angle (\( \theta \)) between the force and displacement vectors.
  • In formula form: \( W = F \cdot d \cdot \cos(\theta) \)
It's important to note that if the force and the movement are aligned, as in this exercise, where the fisherman pulls in line with the net, the angle\( \theta \) is zero, and\( \cos(0) = 1 \).This greatly simplifies the calculation, making it straightforward to compute the work done.
Physics Problems
Physics problems like the one here provide an excellent opportunity to practice applying mathematical formulas in real-world scenarios. Solving physics exercises usually requires identifying what you know,
  • the force exerted
  • the distance involved
  • any relevant angles
and applying the proper formula.
In our example of the fisherman, the challenge is to find how much work is done when pulling in the line. By identifying the known values (force, displacement, and alignment of direction), we use the work formula to solve the problem. This step-by-step approach is crucial as it ensures no aspect of the problem is overlooked.
Force and Displacement
Force and displacement are central concepts in solving physics problems, especially when calculating work. Force is the push or pull on an object, measured in newtons (N) in the SI system. Displacement, on the other hand, measures how far an object moves, measured in meters (m).
In our fishing problem:
  • The fisherman applies a force of \( 25.0 \, \mathrm{N} \)
  • Reels in a line of \( 12.0 \, \mathrm{m} \)
To calculate the work done, identify that both force and displacement are in the same direction. Thus, the angle \( \theta = 0 \), simplifying our computation as the force fully contributes to the work done on the fish.
High School Physics
High school physics often introduces students to the basics of mechanics, including work and energy. The principles learned, such as understanding forces and how they cause movement, set a foundation for exploring more complex physical concepts.
One of the foundational lessons is grasping that work involves not just any force, but the component of the force in the direction of movement. This highlights the importance of relating abstract physics concepts to things we can observe and measure, like the fishing problem.
  • It emphasizes how theory connects to real-world applications
  • Builds critical thinking skills
  • Enhances mathematical aptitude through problem-solving
By consistently applying these principles, students gain a deeper comprehension of how the physical world operates.

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Most popular questions from this chapter

The total height of Yosemite Falls is \(2425 \mathrm{ft}\). (a) How many more joules of gravitational potential energy are there for each kilogram of water at the top of this waterfall compared with each kilogram of water at the foot of the falls? (b) Find the kinetic energy and speed of each kilogram of water as it reaches the base of the waterfall, assuming that there are no losses due to friction with the air or rocks and that the mass of water had negligible vertical speed at the top. How fast (in \(\mathrm{m} / \mathrm{s}\) and \(\mathrm{mph}\) ) would a \(70 \mathrm{~kg}\) person have to run to have that much kinetic energy? (c) How high would Yosemite Falls have to be so that each kilogram of water at the base had twice the kinetic energy you found in part (b); twice the speed you found in part (b)?

You throw a \(20 \mathrm{~N}\) rock into the air from ground level and observe that, when it is \(15.0 \mathrm{~m}\) high, it is traveling upward at \(25.0 \mathrm{~m} / \mathrm{s}\). Use the work-energy theorem to find (a) the rock's speed just as it left the ground and (b) the maximum height the rock will reach.

Marbles of mass \(m\) are thrown from the edge of a vertical cliff of height \(h\) at speed \(v_{0} .\) Neglecting air resistance, how fast (in terms of \(m, h,\) and \(v_{0}\) ) will these marbles be moving when they reach the bottom of the cliff if they are thrown (a) straight up, (b) straight down, or (c) horizontally away from the cliff? Will the final velocity vectors of the marbles be the same or different for each case?

A block of ice with mass \(2.00 \mathrm{~kg}\) slides \(0.750 \mathrm{~m}\) down an inclined plane that slopes downward at an angle of \(36.9^{\circ}\) below the horizontal. If the block of ice starts from rest, what is its final speed? You can ignore friction.

A constant horizontal pull of \(8.50 \mathrm{~N}\) drags a box along a horizontal floor through a distance of \(17.4 \mathrm{~m}\). (a) How much work does the pull do on the box? (b) Suppose that the same pull is exerted at an angle above the horizontal. If this pull now does \(65.0 \mathrm{~J}\) of work on the box while pulling it through the same distance, what angle does the force make with the horizontal?
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