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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) Work done: 105 J. (b) Increase in thermal energy of floor: 30.56 J. (c) Increase in kinetic energy: 34.44 J.

Step by step solution

01

Calculate the work done by the applied force

To find the work done by the applied force, we use the formula for work: \( W = F \cdot d \cdot \cos(\theta) \) where \( F \) is the force, \( d \) is the displacement, and \( \theta \) is the angle between the force and displacement vectors. Here, \( \theta = 0 \) because the force is horizontal and the displacement is in the same direction.Given: \[ F = 35.0 \text{ N}, \quad d = 3.00 \text{ m}, \quad \cos(0) = 1 \]So, the work done is:\[ W = 35.0 \text{ N} \times 3.00 \text{ m} \times 1 = 105 \text{ J} \]
02

Calculate the work done by friction

First, identify the frictional force using the formula: \( f_k = \mu_k \cdot N \), where \( \mu_k \) is the coefficient of kinetic friction, and \( N \) is the normal force. Since there is no vertical motion, \( N = mg \), where \( m = 4.00 \text{ kg} \) and \( g = 9.8 \text{ m/s}^2 \).\[ f_k = 0.600 \times 4.00 \text{ kg} \times 9.8 \text{ m/s}^2 = 23.52 \text{ N} \]The work done by friction, which acts opposite to the displacement, is:\[ W_f = f_k \cdot d \cdot \cos(180^\circ) = -23.52 \text{ N} \times 3.00 \text{ m} \]\[ W_f = -70.56 \text{ J} \]
03

Calculate the increase in kinetic energy of the block

Using the work-energy principle, the net work done on the block is equal to the change in kinetic energy:\[ \Delta K = W_\text{applied} + W_f \]From Step 1, \( W_\text{applied} = 105 \text{ J} \) and from Step 2, \( W_f = -70.56 \text{ J} \).So,\[ \Delta K = 105 \text{ J} - 70.56 \text{ J} = 34.44 \text{ J} \]
04

Calculate the increase in thermal energy of the floor

Given in the problem, the thermal energy of the block increases by \( 40.0 \text{ J} \). The total increase in thermal energy (block plus floor) due to friction must equal the negative of the work done by friction because the lost mechanical energy is converted to thermal energy.This thermal energy increase is:\[ \text{Total Thermal Energy} = -W_f = 70.56 \text{ J} \]Therefore, the increase in thermal energy of the floor is the difference:\[ 70.56 \text{ J} - 40.0 \text{ J} = 30.56 \text{ J} \]

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Key Concepts

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

Kinetic Friction
Kinetic friction is a force that acts between moving surfaces. Let's think about what happens when a block is sliding on a surface. As the block moves, it experiences a resistance due to kinetic friction which acts in the opposite direction to the movement. This force tries to stop the block from moving further.
The magnitude of this frictional force depends on two factors:
  • The coefficient of kinetic friction (\( \mu_k \)) which is a unique value for every pair of surfaces in contact.
  • The normal force (N), which is usually the weight of the object if there's no vertical acceleration, given by \( N = mg \), where m is mass and g is gravitational acceleration.
In our exercise, we calculated this force as \( f_k = 0.600 \times (4.00\, \text{kg} \times 9.8 \text{\,m/s}^2) \) resulting in a force of 23.52 N. This frictional force does negative work on the block, opposing its motion and transforming kinetic energy into thermal energy.
Work-Energy Principle
The work-energy principle is a very important concept in physics that connects the work done by forces to a change in kinetic energy. It states that the total work done on an object is equal to the change in its kinetic energy.
This principle is expressed by the equation:\[ \Delta K = W_{\text{applied}} + W_f \]Where \( \Delta K \) is the change in kinetic energy, \( W_{\text{applied}} \) is the work done by any applied forces, and \( W_f \) is work done by friction or other forces opposing the movement.
In our specific case, the work done by the 35 N force is positive, causing an increase in the block's kinetic energy, calculated as 105 J. Meanwhile, friction does -70.56 J of work. Combining these effects shows an overall increase in kinetic energy of 34.44 J. Hence, the block speeds up by this amount of energy.
Thermal Energy
As objects move against surfaces, some of their kinetic energy is lost to thermal energy due to friction. In simpler terms, when the surfaces slide against each other, they heat up. The lost mechanical energy is transformed into thermal energy, which is an important part of energy conservation.
In our problem, we know friction did \(-70.56\, \text{J} \) of work, implying that this amount was converted to thermal energy. The problem states the block gains 40 J of thermal energy during its slide.
The rest of the energy must have been transferred to the floor because the surfaces interact dynamically and energy is not lost, only transformed or transferred.
Using the data provided, we determine the floor's share of thermal energy increase as: \( 70.56 \text{ J} - 40 \text{ J} = 30.56 \text{ J} \). This conservation and transformation of energy reflect the physical law that energy cannot be created or destroyed, only converted from one form to another.

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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?

A skier weighing \(600 \mathrm{~N}\) goes over a frictionless circular hill of radius \(R=20 \mathrm{~m}\) (Fig. 8-62). Assume that the effects of air resistance on the skier are negligible. As she comes up the hill, her speed is \(8.0 \mathrm{~m} / \mathrm{s}\) at point \(B\), at angle \(\theta=20^{\circ}\). (a) What is her speed at the hilltop (point \(A\) ) if she coasts without using her poles? (b) What minimum speed can she have at \(B\) and still coast to the hilltop? (c) Do the answers to these two questions increase, decrease, or remain the same if the skier weighs \(700 \mathrm{~N}\) instead of \(600 \mathrm{~N} ?\)

A rope is used to pull a \(3.57 \mathrm{~kg}\) block at constant speed \(4.06 \mathrm{~m}\) along a horizontal floor. The force on the block from the rope is \(7.68 \mathrm{~N}\) and directed \(15.0^{\circ}\) above the horizontal. What are (a) the work done by the rope's force, (b) the increase in thermal energy of the block-floor system, and (c) the coefficient of kinetic friction between the block and floor?

The summit of Mount Everest is \(8850 \mathrm{~m}\) above sea level. (a) How much energy would a \(90 \mathrm{~kg}\) climber expend against the gravitational force on him in climbing to the summit from sea level? (b) How many candy bars, at \(1.25 \mathrm{MJ}\) per bar, would supply an energy equivalent to this? Your answer should suggest that work done against the gravitational force is a very small part of the energy expended in climbing a mountain.

A 60 kg skier starts from rest at height H 20 m above the end of a ski-jump ramp (Fig. 8-37) and leaves the ramp at angle u 28. Neglect the effects of air resistance and assume the ramp is frictionless. (a) What is the maximum height h of his jump above the end of the ramp? (b) If he increased his weight by putting on a backpack, would h then be greater, less, or the same?

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