/*! 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 44 A horizontal force of magnitude ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 44

A horizontal force of magnitude \(35.0 \mathrm{~N}\) pushes a block of mass \(4.00 \mathrm{~kg}\) across a floor where the coefficient of kinetic friction is \(0.600 .\) (a) How much work is done by that applied force on the block- floor system when the block slides through a displacement of \(3.00 \mathrm{~m}\) across the floor? (b) During that displacement, the thermal energy of the block increases by \(40.0 \mathrm{~J}\). What is the increase in thermal energy of the floor? (c) What is the increase in the kinetic energy of the block?

Short Answer

Expert verified
(a) 105 J, (b) 30.56 J, (c) 34.44 J.

Step by step solution

01

Calculate the work done by the applied force

To calculate the work done by the applied force, use the formula for work: \[ W = F \cdot d \cdot \cos(\theta), \]where \( F \) is the force applied (35.0 N), \( d \) is the displacement (3.00 m), and \( \theta \) is the angle between the force and displacement direction. Since the force is horizontal and the displacement is also horizontal, \( \theta = 0 \degree \), and \( \cos(0\degree) = 1\). Therefore, \[ W = 35.0 \times 3.00 \times 1 = 105.0 \, \text{J}. \]
02

Calculate the frictional force

The frictional force can be calculated using:\[ f_k = \mu_k \cdot N. \]Here, \( \mu_k \) is the coefficient of kinetic friction (0.600), and \( N \) is the normal force. Since there is no vertical motion, the normal force \( N = mg = 4.00 \times 9.8 = 39.2 \, \text{N}. \)Substitute to find the frictional force:\[ f_k = 0.600 \times 39.2 = 23.52 \, \text{N}. \]
03

Calculate the work done by the frictional force

The work done by the frictional force is given by:\[ W_{\text{friction}} = f_k \cdot d \cdot \cos(180\degree), \]since the frictional force is opposite to the displacement. \( \cos(180\degree) = -1 \), hence:\[ W_{\text{friction}} = 23.52 \times 3.00 \times (-1) = -70.56 \, \text{J}. \]
04

Determine increase in thermal energy of the floor

The work done by friction contributes to the increase in thermal energy of both the block and the floor. Given the block's thermal energy increase is 40.0 J, the remainder of the frictional work increases the floor's thermal energy. Thus:\[ \text{Increase in thermal energy of floor} = 70.56 \, \text{J} - 40.0 \, \text{J} = 30.56 \, \text{J}. \]
05

Calculate the increase in kinetic energy of the block

The increase in kinetic energy of the block can be obtained by subtracting the work done by friction from the work done by the applied force:\[ \Delta KE = W_{\text{applied}} + W_{\text{friction}}. \]Substituting the values gives:\[ \Delta KE = 105.0 + (-70.56) = 34.44 \, \text{J}. \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. In the context of a block sliding across a floor, it's crucial to understand how kinetic energy changes as forces act upon the block. The formula to calculate kinetic energy is given by:
  • \( KE = \frac{1}{2}mv^2 \)
where \( m \) represents mass and \( v \) is velocity.

When a force is applied to an object, causing it to move, work is done, which can translate into an increase in the object's kinetic energy. In the given exercise, the block's kinetic energy increases as it's pushed across the floor, demonstrating energy transfer through work.

The increase in kinetic energy can be calculated as the net work done on the block by various forces like the applied force and frictional force. Here, after accounting for the opposing work of friction, the block's kinetic energy increased by \(34.44 \, \text{J}\). This highlights how crucial kinetic energy concepts are in understanding motion and force interactions.
Thermal Energy
Thermal energy refers to the internal energy present in a system due to its temperature. In this exercise, both the block and the floor experience an increase in thermal energy as the block is pushed across the surface.

The frictional force between the block and the floor generates thermal energy because friction resists motion by converting kinetic energy into heat. This comes into play every time two surfaces slide across each other.
  • In this situation, the block increases its thermal energy by \(40.0 \, \text{J}\).
  • The remaining thermal energy from friction dissipates into the floor, calculated as \(30.56 \, \text{J}\).
This transfer of energy underscores the principle of energy conservation where energy is neither created nor destroyed, but rather transformed from one form to another, in this case from kinetic to thermal.

Understanding this concept is essential in applications where managing heat due to friction is critical, such as in mechanical systems and material engineering.
Frictional Force
Frictional force is a force exerted by a surface to oppose the motion of an object moving across it. It is dependent on two main factors: the nature of the surfaces and the normal force.

The formula for kinetic frictional force is:
  • \( f_k = \mu_k \cdot N \)
where \( \mu_k \) is the coefficient of kinetic friction and \( N \) is the normal force, which is the perpendicular force the object exerts on the surface.

For the exercise, the calculated frictional force was \(23.52 \, \text{N}\), acting opposite to the block's direction of motion. This force performs negative work on the block, meaning it removes energy from the block's kinetic energy and converts it into thermal energy.
  • Frictional work results in a \(-70.56 \, \text{J}\) change, demonstrating energy transformation due to friction.
Recognizing how frictional force operates is vital because it influences energy dynamics in countless daily and industrial contexts.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A collie drags its bed box across a floor by applying a horizontal force of \(8.0 \mathrm{~N}\). The kinetic frictional force acting on the box has magnitude \(5.0 \mathrm{~N}\). As the box is dragged through \(0.70 \mathrm{~m}\) along the way, what are (a) the work done by the collie's applied force and (b) the increase in thermal energy of the bed and floor?

The only force acting on a particle is conservative force \(\vec{F}\). If the particle is at point \(A,\) the potential energy of the system associated with \(\vec{F}\) and the particle is \(40 \mathrm{~J}\). If the particle moves from point \(A\) to point \(B,\) the work done on the particle by \(\vec{F}\) is \(+25 \mathrm{~J}\). What is the potential energy of the system with the particle at \(B ?\)

A volcanic ash flow is moving across horizontal ground when it encounters a \(10^{\circ}\) upslope. The front of the flow then travels \(920 \mathrm{~m}\) up the slope before stopping. Assume that the gases entrapped in the flow lift the flow and thus make the frictional force from the ground negligible; assume also that the mechanical energy of the front of the flow is conserved. What was the initial speed of the front of the flow?

A \(9.40 \mathrm{~kg}\) projectile is fired vertically upward. Air drag decreases the mechanical energy of the projectile Earth system by \(68.0 \mathrm{~kJ}\) during the projectile's ascent. How much higher would the projectile have gone were air drag negligible?

A \(25 \mathrm{~kg}\) bear slides, from rest, \(12 \mathrm{~m}\) down a lodgepole pine tree, moving with a speed of \(5.6 \mathrm{~m} / \mathrm{s}\) just before hitting the ground. (a) What change occurs in the gravitational potential energy of the bear-Earth system during the slide? (b) What is the kinetic energy of the bear just before hitting the ground? (c) What is the average frictional force that acts on the sliding bear?
See all solutions

Recommended explanations on Physics Textbooks

Particle Model of Matter

Read Explanation

Space Physics

Read Explanation

Engineering Physics

Read Explanation

Measurements

Read Explanation

Thermodynamics

Read Explanation

Electromagnetism

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.