/*! 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 30 An object moves \(2.50 \mathrm{~... [FREE SOLUTION] | 91Ó°ÊÓ

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An object moves \(2.50 \mathrm{~m}\) in the \(+x\) -direction under the influence of a \(125-\mathrm{N}\) force directed \(50^{\circ}\) above the \(x\) -axis. Find the work done on the object.

Short Answer

Expert verified
The work done on the object is approximately 200.88 J.

Step by step solution

01

Understand Work Formulation

The work done by a force on an object is given by the formula: \[ W = Fd\cos(\theta) \]where \(W\) is the work done, \(F\) is the magnitude of the force, \(d\) is the distance the object moves, and \(\theta\) is the angle between the force and the direction of motion.
02

Identify Given Variables

From the exercise, the given quantities are:- Magnitude of force, \( F = 125 \mathrm{~N} \)- Distance moved, \( d = 2.50 \mathrm{~m} \)- Angle, \( \theta = 50^{\circ} \) above the \( x \)-axis.
03

Substitute Values into the Work Formula

Substitute the given values into the work formula:\[ W = 125 \mathrm{~N} \times 2.50 \mathrm{~m} \times \cos(50^{\circ}) \]
04

Calculate the Cosine Value

Calculate \( \cos(50^{\circ}) \) using a calculator. \[ \cos(50^{\circ}) \approx 0.6428 \]
05

Compute the Work Done

Using the calculated cosine value,\[ W = 125 \times 2.50 \times 0.6428 = 200.88 \]Thus, the work done on the object is approximately \( 200.88 \mathrm{~J} \).

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Key Concepts

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

Work Done Formula
When we talk about work in physics, it’s about more than getting a job done. In scientific terms, work happens when a force is applied to an object and it moves a certain distance in the direction of the force. The formula to calculate this is:
  • \( W = Fd\cos(\theta) \)
Here’s a break down of each part of the formula:
  • \( W \) represents the work done, usually measured in joules (J).
  • \( F \) is the force applied to the object in newtons (N).
  • \( d \) is the distance that the object has been moved in meters (m).
  • \( \theta \) is the angle between the direction of the applied force and the direction of the object's motion.
Using this formula, we can see that work is maximized when the direction of the force is precisely in line with the direction of movement (\( \theta = 0^{\circ} \)). Conversely, if the force is perpendicular to the movement (\( \theta = 90^{\circ} \)), no work is done because the cosine of 90 degrees is zero. This formula is pivotal in understanding how different forces affect the motion of an object.
Force and Motion
Forces play a crucial role in initiating an object's motion or altering its velocity. In our everyday life, we encounter forces such as the gravitational pull, friction, and applied force—all responsible for the various movements we observe.
An unbalanced force acting on an object will change the object's state of motion according to Newton's First Law. This is where understanding the direction and magnitude of forces becomes essential. When a force is applied at an angle, its effect isn’t just straightforward. Thus, in the given problem, the force acting at an angle causes a component of the force to influence the motion along the x-direction, while the other component may affect motion vertically. It is this horizontal component of the force that becomes part of the work calculation.
The relationship between force and motion can be complex, requiring tools like vector resolutions and understanding trigonometric functions to break down the problem into more manageable components. By resolving the force vectors into perpendicular components, it becomes easier to predict how they will influence motion in different directions.
Angle of Inclination
In the context of work and energy, the angle of inclination refers to the angle \( \theta \) at which the force is applied in relation to the direction of motion. This angle significantly alters how effectively the force contributes to doing work on an object.
The angle of inclination determines the fraction of the total force that contributes to moving an object along the intended direction. When the force is applied directly in the line of motion (\(0^{\circ}\)), the whole force aids the work; however, as this angle increases towards \(90^{\circ}\), the effective component decreases because the force becomes ever more perpendicular. This results in less work being done.
  • To find the effective component of force, we use the cosine of the angle \( \theta \):
\[ F_{\text{effective}} = F \cos(\theta) \]
This makes understanding angles vital, especially in settings involving inclined planes or forces directed off the main axis of movement. In practical terms, this means that tasks such as pushing a sled up a hill or pumping water uphill involve analyzing how much of the initial force is channeled effectively along the desired direction.

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Most popular questions from this chapter

\(\mathrm{A}\) force \(\bar{F}=13 \mathrm{~N} \hat{\imath}+13 \mathrm{~N} \hat{\jmath}\) acts on a hockey puck. Determine the work done if the force results in the puck's displacement by \(4.2 \mathrm{~m}\) in the \(+x\) -direction and \(2.1 \mathrm{~m}\) in the \(-y\) -direction.

A spring with \(k=25.0 \mathrm{~N} / \mathrm{m}\) is oriented vertically with one end fixed to the ground. A \(0.100-\mathrm{kg}\) mass on top of the spring compresses it. Find the spring's maximum compression in each of these cases: (a) You hold the mass while you gently compress the spring, and when you release the mass it sits at rest atop the spring. (b) You place the mass on the uncompressed spring and release it. (c) You drop the mass from \(10.0 \mathrm{~cm}\) above the spring.

A \(75-\mathrm{g}\) toy rocket is launched straight up from the ground at \(19 \mathrm{~m} / \mathrm{s}\). (a) What's its kinetic energy? (b) Find the work done by gravity and the rocket's new kinetic energy after it has risen \(10 \mathrm{~m}\) (c) Use your answer to part (b) to find the rocket's speed at \(10 \mathrm{~m}\).

A Hooke's law spring hangs vertically with the top end fixed. Attaching a \(0.150-\mathrm{kg}\) mass to the bottom end stretches the spring \(0.125 \mathrm{~m}\). (a) Find the spring constant. (b) What will be the total stretch if a \(1.00-\mathrm{kg}\) mass is hung from the spring?

A mass hanging from a vertical spring has gravitational potential energy, and the spring has elastic potential energy. (a) Determine how far the spring \((k=16 \mathrm{~N} / \mathrm{m})\) stretches when a 100 -gram mass is hung from it and allowed to come to rest. (b) If the mass is pulled down \(3 \mathrm{~cm}\) further, determine the change in each type of potential energy.

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