/*! 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 block of mass \(2.50 \mathrm{~... [FREE SOLUTION] | 91Ó°ÊÓ

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A block of mass \(2.50 \mathrm{~kg}\) is pushed \(2.20 \mathrm{~m}\) along a frictionless horizontal table by a constant \(16.0-\mathrm{N}\) force directed \(25.0^{\circ}\) below the horizontal. Determine the work done by (a) the applied force, (b) the normal force exerted by the table, (c) the force of gravity, and (d) the net force on the block.

Short Answer

Expert verified
The work done by (a) the applied force is \(16.0 N \cdot 2.20 m \cdot cos(25.0^{\circ})\) joules, (b) the normal force is zero joules, (c) the force of gravity is also zero joules, and (d) the net force is the same as the applied force, \(16.0 N \cdot 2.20 m \cdot cos(25.0^{\circ})\) joules.

Step by step solution

01

Calculate work done by the applied force.

The applied force and the displacement are not in the same direction. They form an angle of \(25.0^{\circ}\) with each other. So, to calculate the work done \(W_A\) by the applied force, we use the formula: \(W_A = F \cdot d \cdot cos(\theta)\), where \(F = 16.0 N\) is the applied force, \(d = 2.20 m\) is the displacement, and \(\theta = 25.0^{\circ}\). Hence, \(W_A = 16.0 N \cdot 2.20 m \cdot cos(25.0^{\circ})\).
02

Calculate work done by the normal force.

The normal force acts perpendicular to the displacement of the block. Therefore, the angle between the normal force and the displacement is \(90^{\circ}\). The Work done by a force is zero if the force is perpendicular to the direction of displacement. Hence, the work done \(W_N\) by the normal force is zero.
03

Calculate work done by gravity.

The force of gravity also acts perpendicular to the displacement of the block. Therefore, as with the normal force, the work done \(W_G\) by gravity is also zero.
04

Calculate work done by the net force.

In this case, the net force on the block is the same as the applied force. Therefore, the work done \(W_Net\) by the net force is the same as the work done by the applied force. Therefore, \(W_Net = W_A\).

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Key Concepts

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

Applied Force
When we talk about the concept of applied force, we're referring to the force that is deliberately applied to an object to make it move or change its motion. In this exercise, the block is pushed with an applied force of 16.0 N. However, this force is not applied directly horizontally. It is angled 25.0 degrees below the horizontal, which affects how we calculate the work done. When calculating the work done by this force, we use the formula:
  • \( W_A = F \times d \times \cos(\theta) \)
This formula accounts for the fact that only the component of the force in the direction of the movement contributes to the work done. Here, \( F \) is the force (16.0 N), \( d \) is the displacement (2.20 m), and \( \theta \) is the angle (25 degrees). The cosine function helps us find the component of the force in the direction of motion, making sure our calculations are accurate. By doing these calculations, students will discover how angles affect the efficiency of force applied in moving objects.
Normal Force
Normal force is the force exerted by a surface perpendicular to the object resting on it. It is the table pushing back up on the block to support it against gravity. In this case, as the block moves across the table, the normal force acts perpendicularly to the block's path of motion. Because normal force is always perpendicular to the object's displacement, the angle between the normal force and displacement is 90 degrees. This is important because when calculating work, the formula uses the cosine of the angle. The cosine of 90 degrees is zero, which means the work done by the normal force in this scenario is zero. Even though the normal force is essential for keeping the block from accelerating towards the ground, it doesn't directly contribute to the block's horizontal movement. This highlights how directionality is a key factor in understanding forces and their effects.
Gravity
Gravity is a constant force acting on any object with mass, directed towards the center of the Earth. For a block on a table, this force pulls it downwards with an intensity equal to its weight: mass multiplied by the gravitational acceleration, which is approximately 9.81 m/s².In this setup, gravity also acts perpendicularly to the direction of the displacement (horizontal movement) of the block. As with the normal force, when considering the work done by gravity, the angle is 90 degrees. Consequently:
  • The work done by gravity is \( W_G = 0 \)
because the gravitational force does not contribute to the horizontal displacement of the block. Examining gravity's role in various scenarios helps students understand how it underpins fundamental principles of physics, such as potential and kinetic energy relationships.
Net Force
Net force refers to the overall force acting on an object when all individual forces are combined. For this block, the key players are the applied force, the normal force, and gravity. Since normal force and gravity do not perform any work in the horizontal movement of the block due to their perpendicular nature, they do not contribute to the net work done on the block.In situations like this, the net force is effectively the same as the applied force when considering work done. We calculate the work done by the net force the same way we do for the applied force since they are aligned:
  • \( W_{Net} = W_A \)
By understanding net force, students learn to evaluate multiple forces acting simultaneously, comprehend their individual contributions, and determine the overall movement effects on an object. This is crucial for solving real-world physics problems where multiple forces complicate simple analyses.

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Most popular questions from this chapter

A \(2.1 \times 10^{3}-\mathrm{kg}\) car starts from rest at the top of a \(5.0-\mathrm{m}-\) long driveway that is inclined at \(20^{\circ}\) with the horizontal. If an average friction force of \(4.0 \times 10^{9} \mathrm{~N}\) impedes the motion, find the speed of the car at the bottom of the driveway.

An older-model car accelerates from 0 to speed \(v\) in \(10 \mathrm{~s}\). A newer, more powerful sports car of the same mass accelerates from 0 to \(2 v\) in the same time period. Assuming the energy coming from the engine appears only as kinetic encrgy of the cars, compare the power of the two cars.

When a \(2.50-\mathrm{kg}\) object is hung vertically on a certain light spring described by Hooke's law, the spring stretches \(2.76 \mathrm{em}\). (a) What is the force constant of the spring? (b) If the \(2.50-\mathrm{kg}\) object is removed, how far will the spring stretch if a \(1.25-\mathrm{kg}\) block is hung on it? (c) How much work must an external agent do to stretch the same spring \(8.00 \mathrm{~cm}\) from its unstretched position?

A \(60.0\) -kg athlete leaps straight up into the air from a trampoline with an initial speed of \(9.0 \mathrm{~m} / \mathrm{s}\). The goal of this problem is to find the maximum height she attains and her speed at half maximum height. (a) What are the interacting objects and how do they interact? (b) Select the height at which the athlete's speed is \(9.0 \mathrm{~m} / \mathrm{s}\) as \(y=0 .\) What is her kinetic energy at this point? What is the gravitational potential energy associated with the athlete? (c) What is her kinetic energy at maximum height? What is the gravitational potential energy associated with the athlete? (d) Write a general equation for energy conservation in this case and solve for the maximum height. Substitute and obtain a numerical answer. (c) Write the general equation for energy conservation and solve for the velocity at half the maximum height. Substitute and obtain a numerical answer.

A daredevil wishes to bungee-jump from a hot-air balloon \(65.0 \mathrm{~m}\) above a carnival midway (Fig. P5.83). He will use a piece of uniform elastic cord tied to a harness around his body to stop his fall at a point \(10.0 \mathrm{~m}\) above the ground. Model his body as a particle and the cord as having negligible mass and a tension force described by Hooke's force law. In a preliminary test, hanging at rest from a \(5.00-\mathrm{m}\) length of the cord, the jumper finds that his body weight stretches it by \(1.50 \mathrm{~m} .\) He will drop from rest at the point where the top end of a longer section of the cord is attached to the stationary balloon. (a) What length of cord should he use? (b) What maximum acceleration will he experience?

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