/*! 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 8 A woman exerts a horizontal forc... [FREE SOLUTION] | 91Ó°ÊÓ

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A woman exerts a horizontal force of 1 pounds on a box as she pushes it up a ramp that is 3 feet long and inclined at an angle of 30 degrees above the horizontal. Find the work done on the box. Work: ________ft-lb

Short Answer

Expert verified
Work: \( \approx 2.60 \) ft-lb

Step by step solution

01

Find the force parallel to the ramp

To find the force parallel to the ramp, we need to determine the component of the horizontal force exerted by the woman in the direction parallel to the ramp. This can be done using trigonometry. Since the force is exerted horizontally, and the angle of inclination of the ramp is given as 30 degrees, we can use the cosine function to find the component of force parallel to the ramp: Force_parallel = Force_horizontal * cos(angle) Force_parallel = 1 lbs * cos(30)
02

Calculate the component of horizontal force in the direction parallel to the ramp

Now, we need to plug the values into the equation to solve for the component of horizontal force in the direction parallel to the ramp. Force_parallel = 1 lbs * cos(30) Force_parallel ≈ 1 lbs * 0.866 Force_parallel ≈ 0.866 lbs
03

Calculating work done on the box

Now that we have the component of force parallel to the ramp, we can calculate the work done on the box using the work formula: Work_done = Force_parallel × distance Plug in the values for Force_parallel and distance (length of the ramp): Work_done = 0.866 lbs × 3 ft
04

Calculate the work done on the box

Calculate the final value of work done on the box by multiplying the force parallel and distance: Work_done = 0.866 lbs × 3 ft Work_done ≈ 2.598 ft-lb Since significant figures are not specified, the value can be represented with 3 significant figures: Work_done ≈ 2.60 ft-lb

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Key Concepts

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

Trigonometry in Physics
Trigonometry plays a crucial role in physics, especially when it comes to analyzing forces and motion in different directions. It's the mathematical tool that enables us to break down forces into perpendicular components, which are easier to work with individually. In the example of the woman pushing a box up a ramp, we had to find the component of force that's actually acting parallel to the incline, because that's the force doing the work of moving the box upwards.

Using trigonometric functions, like the cosine for angles less than 90 degrees, helps us calculate this parallel component. Here's the key formula used in the exercise:
\[ \text{Force}_{\text{parallel}} = \text{Force}_{\text{horizontal}} \times \cos(\text{angle}) \]
Remember, this formula only works because the initial force is horizontal. If the applied force had been at a different angle, we would've needed to adjust the trigonometric function accordingly. When working with angles and forces, a clear understanding of trigonometry is essential to solve the problem correctly.
Inclined Plane
An inclined plane is a flat surface that is tilted at an angle to the horizontal. This seemingly simple machine eases the workload when lifting or moving objects upwards against gravity. In physics problems, an inclined plane requires you to consider both the gravitational force pulling the object down the slope and any applied forces moving the object up.

When considering work performed on an object on an inclined plane, only the force component parallel to the plane—causing the object to slide or move along the plane—contributes to the work. The perpendicular component (against the plane's surface) is typically responsible for frictional forces but does not contribute to the work done in moving the object along the plane.
Force Parallel Calculation
To calculate the force parallel to an inclined plane, as in our box-pushing example, we first identify the angle of the plane and the direction of the applied force. We then find the force component that is in line with the plane's surface. This is critical, because it's this force that actually moves the object along the plane.

The trigonometric function used depends on the angle at which the force is applied. If the force is applied perpendicular to the base of the incline, the sine function would be used. However, in this exercise, the force is applied horizontally, so the cosine function is relevant.
\[ \text{Force}_{\text{parallel}} = \text{Force}_{\text{horizontal}} \times \cos(\text{angle of incline}) \]
Thus, by calculating the horizontal force's parallel component, we can determine just how much of that force is effectively used to push the object up the incline.
Work-Energy Principle
The work-energy principle is a fundamental concept in physics that describes the relationship between the work done on an object and its energy. When you do work on an object, you're essentially transferring energy to it. In the context of the inclined plane problem, we calculate work done using the formula:
\[ \text{Work}_{\text{done}} = \text{Force}_{\text{parallel}} \times \text{distance} \]
Here, distance is the length of the path over which the force is applied—in this case, the length of the ramp. This work done on the box turns into kinetic energy if the box speeds up, or potential energy if its vertical position within a gravitational field changes. These concepts are intertwined, showing that physics problems are a blend of abstract principles and concrete calculations.

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Most popular questions from this chapter

You are looking down at a map. A vector \(\mathbf{u}\) with \(|\mathbf{u}|=7\) points north and a vector \(\mathbf{v}\) with \(|\mathbf{v}|=6\) points northeast. The crossproduct \(\mathbf{u} \times \mathbf{v}\) points: A) south B) northwest C) up D) down Please enter the letter of the correct answer: _______ The magnitude \(|\mathbf{u} \times \mathbf{v}|=\) _________

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