/*! 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 21 A sled with mass 8.00 \(\mathrm{... [FREE SOLUTION] | 91Ó°ÊÓ

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A sled with mass 8.00 \(\mathrm{kg}\) moves in a straight line on a frictionless horizontal surface. At one point in its path, its speed is 4.00 \(\mathrm{m} / \mathrm{s}\) : after it has traveled 2.50 \(\mathrm{m}\) beyond this point, its speed is 6.00 \(\mathrm{m} / \mathrm{s}\) . Use the work-energy theorem to find the force acting on the sled, assuming that this force is constant and that it acts in the direction of the sled's motion.

Short Answer

Expert verified
The constant force acting on the sled is 32.00 N.

Step by step solution

01

Understanding the Work-Energy Theorem

The work-energy theorem states that the work done by the forces on an object equals the change in its kinetic energy. Mathematically, this is expressed as \( W = \Delta KE \). In this problem, the force is constant and acts along the direction of the sled's motion, making it easy to calculate the work done.
02

Calculating Initial and Final Kinetic Energy

Kinetic energy (\( KE \)) is given by the formula \( KE = \frac{1}{2}mv^2 \). The initial kinetic energy (\( KE_i \)) with speed 4.00 m/s is \( \frac{1}{2} \times 8.00 \times (4.00)^2 = 64.00 \) J. The final kinetic energy (\( KE_f \)) with speed 6.00 m/s is \( \frac{1}{2} \times 8.00 \times (6.00)^2 = 144.00 \) J.
03

Applying the Work-Energy Theorem

From the work-energy theorem, \( W = KE_f - KE_i \). Substituting the calculated kinetic energies, \( W = 144.00 - 64.00 = 80.00 \) J. This work is done by the force over a distance of 2.50 m.
04

Calculating the Force

Work is also calculated by \( W = F \times d \), where \( F \) is the force and \( d \) is the distance. Using \( W = 80.00 \) J and \( d = 2.50 \) m, we rearrange to find \( F = \frac{W}{d} = \frac{80.00}{2.50} = 32.00 \) N. Thus, the constant force acting on the sled is 32.00 N.

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Key Concepts

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

Kinetic Energy
Kinetic energy is a fundamental concept in physics that deals with the energy an object possesses due to its motion. Whenever an object moves, it carries kinetic energy, which is given by the formula: \[ KE = \frac{1}{2}mv^2 \]where \(m\) represents the mass of the object and \(v\) is its speed. This means that both the mass and the speed of an object contribute to its kinetic energy. The faster an object moves, the more kinetic energy it has. Similarly, the larger its mass, the more kinetic energy it carries.
In the context of our sled example, understanding kinetic energy helps us see how the sled's speed changes as it moves along the horizontal surface. When the sled's speed increases from 4.00 m/s to 6.00 m/s, its kinetic energy increases from 64.00 J to 144.00 J, indicating that additional energy is imparted to it during its motion. This change plays a crucial role in determining the amount of work done on the sled by external forces.
Constant Force
When a force acts on an object without changing direction or magnitude, it is considered a constant force. In physics, especially when using the work-energy theorem, constant forces simplify calculations since they provide a consistent value that can be used across the problem.
The sled in our problem is influenced by such a constant force. The work done on the sled by this force leads to a change in its kinetic energy. Because the force is constant, we use the formula:\[ W = F \times d \]Here, \(W\) is the work done by the force, \(F\) is the constant force, and \(d\) is the distance traveled by the object.Understanding constant forces allows us to solve for the force acting on the sled. By knowing that the work done is 80.00 J over a distance of 2.50 m, we can deduce that the constant force exerted is 32.00 N. This kind of calculation is straightforward due to the predictability and simplicity associated with constant forces.
Physics Problem Solving
Physics problem solving often involves applying fundamental principles to real-world situations. One of the essential skills is understanding and utilizing the work-energy theorem to bridge the gap between an object's energy states and the forces acting upon it.
To effectively solve physics problems like the sled example, follow these steps:
  • Identify the principle that applies, such as the work-energy theorem.
  • Calculate the initial and final states of the object, in this case, its kinetic energy.
  • Relate these values through known formulas to find missing quantities, like the force.
In our sled example, we began by calculating the initial and final kinetic energy, then related these to work done using the work-energy theorem. This helped us derive the force acting on the sled.
Being systematic and methodical in your calculations ensures accurate problem solving. By decomposing complex problems into simpler steps and using established formulas, physics problems become much easier to tackle.

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Most popular questions from this chapter

Crash Barrier. A student proposes a design for an auto- mobile crash barrier in which a \(1700-\mathrm{kg}\) sport utility vehicle moving at 20.0 \(\mathrm{m} / \mathrm{s}\) crashes into a spring of negligible mass that slows it to a stop. So that the passengers are not injured, the acceleration of the vehicle as it slows can be no greater than 5.00\(g\) . (a) Find the required spring constant \(k\) , and find the distance the spring will compress in slowing the vehicle to a stop. In your calculation, disregard any deformation or crumpling of the vehicle and the friction between the vehicle and the ground. (b) What disadvantages are there to this design?

A little red wagon with mass 7.00 \(\mathrm{kg}\) moves in a straight line on a frictionless horizontal surface. It has an initial speed of 4.00 \(\mathrm{m} / \mathrm{s}\) and then is pushed 3.0 \(\mathrm{m}\) in the direction of the initial velocity by a force with a magnitude of 10.0 \(\mathrm{N}\) . (a) Use the work- energy theorem to calculate the wagon's final speed. (b) Calculate the acceleration produced by the force. Use this acceleration in the kinematic relationships of Chapter 2 to calculate the wagon's final speed. Compare this result to that calculated in part (a).

It takes a force of 53 \(\mathrm{kN}\) on the lead car of a 16 -car passenger train with mass \(9.1 \times 10^{5} \mathrm{kg}\) to pull it at a constant 45 \(\mathrm{m} / \mathrm{s}\) \((101 \mathrm{mi} / \mathrm{h})\) on level tracks. (a) What power must the locomotive \((101 \mathrm{mide} \text { to the lead car? (b) How much more power to the lead car }\) than calculated in part (a) would be needed to give the train an up a 1.5\(\%\) grade (slope angle \(\alpha=\arctan 0.015\) ) at a constant 45 \(\mathrm{m} / \mathrm{s} ?\)

The aircraft carrier John \(F .\) Kennedy has mass \(7.4 \times 10^{7} \mathrm{kg}\). When its engines are developing their full power of \(280,000 \mathrm{hp}\), the John \(F\) . Kennedy travels at its top speed of 35 knots \((65 \mathrm{km} / \mathrm{h})\). If 70\(\%\) of the power output of the engines is applied to pushing the ship through the water, what is the magnitude of the force of water resistance that opposes the carrier's motion at this speed?

Rotating Bar. A thin, uniform \(12.0-\mathrm{kg}\) bar that is 2.00 \(\mathrm{m}\) long rotates uniformly about a pivot at one end, making 5.00 complete revolutions every 3.00 seconds. What is the kinetic energy of this bar? (Hint: Different points in the bar have different speeds. Break the bar up into infinitesimal segments of mass \(d m\) and integrate to add up the kinetic energy of all these segments.)

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