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Supppose you pull a sled with a rope that makes an angle of \(30.0^{\circ}\) to the horizontal. How much work do you do if you pull with \(25.0 \mathrm{~N}\) of force and the sled moves \(25.0 \mathrm{~m} ?\)

Short Answer

Expert verified
Question: Calculate the work done by a person pulling a sled with a force of 25 N at an angle of 30 degrees above the horizontal for a distance of 25 meters. Answer: The work done in pulling the sled is approximately 541.2 joules.

Step by step solution

01

Convert the angle to radians

Converting the given angle from degrees to radians using the formula radians=\(\frac{(degrees)(\pi)}{180}\): Angle = \(\frac{30.0 \times \pi}{180} = \frac{\pi}{6}\) radians.
02

Calculate the horizontal component of the force

The horizontal component of the force can be calculated by multiplying the force with the cosine of the angle as follows: Horizontal Force = Force × cos(angle) Horizontal Force = \(25.0 N \times \cos(\frac{\pi}{6})\)
03

Calculate the work done by the horizontal component of the force

The work done can be calculated by multiplying the horizontal component of the force with the distance as follows: Work = Horizontal Force × Distance Work = \((25.0 N \times \cos(\frac{\pi}{6})) \times 25.0 m\)
04

Calculate the final result

Evaluate the expression to find the work done: Work = \((25.0 N \times \cos(\frac{\pi}{6})) \times 25.0 m \approx 541.2 J\) The work done in pulling the sled is approximately 541.2 joules.

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Key Concepts

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

Calculating Work in Physics
The concept of work in physics is fundamental to understanding how energy is transferred by a force moving an object. Work is calculated as the product of the force applied to an object and the distance over which it is applied. But it's not as simple as multiplying any force by any distance. The force must be in the direction of the movement. The equation to remember is: \[ W = F \times d \times \cos(\theta) \], where \( W \) is work, \( F \) is the magnitude of the force, \( d \) is the distance moved by the object, and \( \theta \) is the angle between the direction of the force and the direction of the movement.

Knowing this equation, we see that only the component of force that acts in the direction of motion does actual work on the object. If the force is perpendicular to the displacement (like lifting a weight vertically while walking forward), the work done in the direction of displacement is zero. While it may seem complex, understanding this formula enables you to solve a wide array of problems in physics.
Horizontal Component of Force
When a force is applied at an angle, not all of that force is effective in moving an object along the horizontal. The horizontal component of the applied force is what does work in moving an object horizontally. The horizontal force is found by multiplying the total force by the cosine of the angle of application: \[ F_{x} = F \times \cos(\theta) \], where \( F_{x} \) is the horizontal force, \( F \) is the applied force, and \( \theta \) is the angle. The angle's cosine effectively 'scales down' the force to its horizontal component.

It's crucial to understand the horizontal component of force because this is precisely the part that does the work when an object moves horizontally. Having the ability to sort which portion of the force is contributing to the work done prevents misconceptions about the efficiency and energy requirements of tasks.
Angle Conversion to Radians
Angles can be measured in degrees or radians, and it's important in physics to be able to convert from one to the other. Radians are a 'natural' unit of angular measure in mathematics and physics because they relate the arc length directly to the radius of a circle. To convert degrees to radians, we use the conversion factor \( \frac{\pi \text{ radians}}{180^\circ} \). The conversion formula is \[ \text{radians} = \frac{(\text{degrees})(\pi)}{180} \].

In physics exercises like calculating work, forces or motions on paths that involve circular arcs, and many other applications, radians are essential. Moreover, trigonometric functions in calculators are often set to use radians, and using the incorrect unit could lead to errors in calculations.
Force-Distance Relationship
In the context of work, the force-distance relationship refers to how the work done is dependent on both the magnitude of the force applied and the distance over which it is applied. The core concept here is that both factors — force and distance — are equally important. An increased force on an object over the same distance results in more work done, and similarly, applying a force over a greater distance also increases the work. The work's dependence on distance is linear, which means that doubling the distance will double the work done, as long as the force remains constant. This relationship is captured in the work equation: \[ W = F \times d \], which simplifies the general expression \( W = F \times d \times \cos(\theta) \) when the force is applied in the direction of motion (\( \theta = 0^\circ \)).

Understanding the force-distance relationship helps in analyzing systems' energy requirements and designing systems in engineering, from simple pulleys to complex machines.

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Most popular questions from this chapter

A father pulls his son, whose mass is \(25.0 \mathrm{~kg}\) and who is sitting on a swing with ropes of length \(3.00 \mathrm{~m}\), backward until the ropes make an angle of \(33.6^{\circ}\) with respect to the vertical. He then releases his son from rest. What is the speed of the son at the bottom of the swinging motion?

Eight books, each \(4.6 \mathrm{~cm}\) thick and of mass \(1.8 \mathrm{~kg}\), lie on a flat table. How much work is required to stack them on top of one another? a) 141 J c) 230 e) 14 b) 23 J d) 0.81

A car of mass \(m\) accelerates from rest along a level straight track, not at constant acceleration but with constant engine power, \(P\). Assume that air resistance is negligible. a) Find the car's velocity as a function of time. b) A second car starts from rest alongside the first car on the same track, but maintains a constant acceleration. Which car takes the initial lead? Does the other car overtake it? If yes, write a formula for the distance from the starting point at which this happens. c) You are in a drag race, on a straight level track, with an opponent whose car maintains a constant acceleration of \(12.0 \mathrm{~m} / \mathrm{s}^{2} .\) Both cars have identical masses of \(1000 . \mathrm{kg} .\) The cars start together from rest. Air resistance is assumed to be negligible. Calculate the minimum power your engine needs for you to win the race, assuming the power output is constant and the distance to the finish line is \(0.250 \mathrm{mi}\)

Calculate the power required to propel a \(1000.0-\mathrm{kg}\) car at \(25.0 \mathrm{~m} / \mathrm{s}\) up a straight slope inclined \(5.0^{\circ}\) above the horizontal. Neglect friction and air resistance.

A horse draws a sled horizontally on snow at constant speed. The horse can produce a power of \(1.060 \mathrm{hp} .\) The coefficient of friction between the sled and the snow is \(0.115,\) and the mass of the sled, including the load, is \(204.7 \mathrm{~kg}\). What is the speed with which the sled moves across the snow?

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