/*! 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 43 A collie drags its bed box acros... [FREE SOLUTION] | 91Ó°ÊÓ

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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?

Short Answer

Expert verified
(a) 5.6 J, (b) 3.5 J.

Step by step solution

01

Understand the Concept of Work Done

Work done by a force is calculated using the formula: \( W = F \cdot d \cdot \cos(\theta) \). Here, \( F \) is the applied force, \( d \) is the displacement, and \( \theta \) is the angle between the force and displacement vectors. In this problem, the force is horizontal and displacement is horizontal, so \( \theta = 0^\circ \) and \( \cos(\theta) = 1 \).
02

Calculate the Work Done by the Applied Force

Substitute the given values into the work done formula: \( W = F \cdot d \cdot \cos(\theta) \) becomes \( W = 8.0 \mathrm{~N} \cdot 0.70 \mathrm{~m} \cdot 1 \). Calculate to get \( W = 5.6 \mathrm{~J} \).
03

Understand the Concept of Thermal Energy Increase

The increase in thermal energy due to kinetic friction is equal to the work done against the frictional force, as there is no change in kinetic energy. This can be calculated using the same work done formula, considering frictional force and displacement.
04

Calculate the Increase in Thermal Energy

Use the formula \( W = F_{\text{friction}} \cdot d \), substituting the values: \( W = 5.0 \mathrm{~N} \cdot 0.70 \mathrm{~m} \). Calculate to get \( W = 3.5 \mathrm{~J} \).

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Key Concepts

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

Kinetic Friction
Kinetic friction acts on objects in motion, like the bed box being dragged by the collie. This force, caused by the interaction of the surfaces of the bed and the floor, resists the motion.

**Characteristics of Kinetic Friction:**
  • It always acts in the direction opposite to the motion.
  • The magnitude is constant for a given pair of surfaces and consistent state of motion.
  • Coefficient of kinetic friction, usually denoted as \( \mu_k \), helps express this force: \( F_{\text{friction}} = \mu_k \cdot N \), where \( N \) is the normal force exerted by a surface.
In our problem, the kinetic friction force is \( 5.0 \mathrm{~N} \), and it's this force that the collie must overcome to move the box. Understanding kinetic friction is crucial in these scenarios, as it absorbs energy that otherwise could cause acceleration, transforming it into thermal energy instead.
Thermal Energy
Thermal energy, often resulting from friction, refers to the internal energy present due to the random motion and collisions of the particles in a substance. When the collie drags its bed box, kinetic friction generates thermal energy that slightly warms the surfaces.

**Key Aspects of Thermal Energy:**
  • Produced from the work done against kinetic friction.
  • Related to the increase in temperature, although this might be noticeable only under certain conditions.
  • Transfer is spontaneous from hot to cold regions, ensuring thermodynamic balance.
In the problem, the increase in thermal energy due to dragging is given by the work done against friction: \( 3.5 \mathrm{~J} \). This means, out of the total energy the collie expends, this much is converted into heat rather than motion, demonstrating energy conservation principles in physics.
Work Done
Work done in physics relates to energy transfer when a force moves an object over a distance. It's a fundamental concept that links forces to energy changes. In our case, the collie applies a horizontal force to drag the bed box, and work done helps quantify the energy transferred through this action.

**Work Done Formula:** Using the formula \( W = F \cdot d \cdot \cos(\theta) \), it represents the work done by the force:
  • \( W \) is the work done, measured in joules.
  • \( F \) is the force applied, \( d \) is the displacement, and \( \theta \) is the angle between the force and displacement.
  • In the horizontal drag scenario, \( \theta \) is \( 0^\circ \), hence \( \cos(0) = 1 \).
For the collie's effort:
\( W = 8.0 \mathrm{~N} \times 0.70 \mathrm{~m} \times 1 = 5.6 \mathrm{~J} \). This correctly indicates the energy imparted into moving the box, reflecting how various physics elements, like forces and motion, integrate into everyday activities.

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Most popular questions from this chapter

In a circus act, a \(60 \mathrm{~kg}\) clown is shot from a cannon with an initial velocity of \(16 \mathrm{~m} / \mathrm{s}\) at some unknown angle above the horizontal. A short time later the clown lands in a net that is \(3.9 \mathrm{~m}\) vertically above the clown's initial position. Disregard air drag. What is the kinetic energy of the clown as he lands in the net?

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