/*! 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 3 A moving 300 -g object slides un... [FREE SOLUTION] | 91Ó°ÊÓ

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A moving 300 -g object slides unpushed \(80 \mathrm{~cm}\) in a straight line along a horizontal tabletop. How much work is done in overcoming friction between the object and the table if the coefficient of kinetic friction is \(0.20\) ? First find the friction force. Since the normal force equals the weight of the object, $$ F_{\mathrm{f}}=\mu_{k} F_{N}=(0.20)(0.300 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=0.588 \mathrm{~N} $$ The work done overcoming friction is \(F_{\mathrm{f}} S \cos \theta\). Here \(\theta\) is the angle between the force and the displacement. Because the friction force is opposite in direction to the displacement, \(\theta=180^{\circ}\). Therefore, Work \(=F_{\mathrm{f}} S \cos 180^{\circ}=(0.588 \mathrm{~N})(0.80 \mathrm{~m})(-1)=-0.47 \mathrm{~J}\) The work is negative because the friction force is oppositely directed to the displacement; it slows the object and it decreases the object's kinetic energy, or more to the point, it opposes the motion.

Short Answer

Expert verified
The work done by friction is -0.47 J.

Step by step solution

01

Identify the Given Values

The object has a mass of 300 g (or 0.300 kg when converted to kg), it moves 0.80 m (80 cm) across the table, and the coefficient of kinetic friction \(\mu_k\) is 0.20.
02

Calculate the Normal Force

The normal force \(F_N\) is equal to the weight of the object. The weight is calculated by \(F_N = m \cdot g = 0.300 \text{ kg} \times 9.81 \text{ m/s}^2 = 2.943 \text{ N}\).
03

Calculate the Frictional Force

The frictional force \(F_f\) is given by the formula \(F_f = \mu_k \cdot F_N = 0.20 \times 2.943 \text{ N} = 0.5886 \text{ N}\).
04

Understand the Work Formula

Work done by a force is calculated as \(W = F \cdot S \cdot \cos \theta\), where \(S\) is the displacement, and \(\theta\) is the angle between the force and the direction of displacement. For friction, \(\theta = 180^{\circ}\), as it acts opposite to displacement.
05

Calculate the Work Done

Plug the values into the work formula. Thus, \(W = F_f \cdot S \cdot \cos 180^{\circ} = 0.5886 \text{ N} \times 0.80 \text{ m} \times (-1) = -0.47088 \text{ J}\).
06

Interpret the Result

The negative sign indicates that the frictional force is opposite to the direction of motion, meaning energy is removed from the object as it slides, slowing it down.

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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 resists the motion of an object sliding across a surface. It comes into play when there is relative motion between the two surfaces in contact. The magnitude of kinetic friction depends on two main factors: the normal force and the coefficient of kinetic friction.
- The coefficient of kinetic friction (\(\mu_k\)) is a dimensionless constant that represents the frictional properties of the two surfaces in contact. Each pair of surfaces has a unique \(\mu_k\) value.- The normal force is the perpendicular force exerted by the surface on the object, and kinetic friction is calculated by multiplying this with \(\mu_k\). So, the kinetic friction formula is: \(F_f = \mu_k \cdot F_N\).
Kinetic friction always opposes the direction of motion, which means it does negative work on the object, reducing its kinetic energy and slowing it down.
Normal Force
The normal force is the force that a surface exerts perpendicularly to an object resting on it. It acts in opposition to the object's weight when the object is on a horizontal plane.
- The normal force is crucial because it determines the frictional force; without it, there would be no frictional resistance.
For an object simply resting on a flat surface without other external forces acting vertically, the normal force is equal in magnitude to the object's weight (mass times gravity). Therefore, \(F_N = m \cdot g\), where \(m\) is the mass and \(g\) is the acceleration due to gravity.
This relationship means that any change in weight (such as adding additional loads on the object) directly affects the normal force, subsequently impacting the kinetic friction experienced by the object.
Negative Work
Negative work occurs when the force exerted on an object is in the opposite direction to the object's displacement. This is a common situation when dealing with frictional forces.
- In physics, work is calculated using the formula \(W = F \cdot S \cdot \cos \theta\), where \(F\) is the force applied, \(S\) is the displacement, and \(\theta\) is the angle between force and displacement directions.
- When \(\theta = 180^{\circ}\), as in the case of frictional forces, \(\cos 180^{\circ} = -1\), hence making the work done negative.
Negative work implies energy is being removed from the system, as seen with the frictional force against a sliding object. This energy is often converted into heat, resulting in the object slowing down.
Frictional Force
Frictional force is a necessary and disruptive force that acts parallel to the surfaces in contact, directly opposing the motion or attempted motion. It can be categorized mainly into two types: static and kinetic friction. Our focus here is on kinetic friction.
- Kinetic friction comes into play when objects are already sliding against each other. It is less than static friction, which acts when the objects are at rest.- To calculate the frictional force, you need the normal force and the coefficient of kinetic friction.
The equation for kinetic frictional force is \(F_f = \mu_k \cdot F_N\), meaning the force depends on the surface texture and the object's weight effectively pressed against the surface.
While frictional force can be troublesome by causing wear and energy loss, it is also essential for daily activities like walking, gripping, and driving.

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Most popular questions from this chapter

The coefficient of sliding friction between a \(900-\mathrm{kg}\) car and the pavement is \(0.80\). If the car is moving at \(25 \mathrm{~m} / \mathrm{s}\) along level pavement when it begins to skid to a stop, how far will it go before coming to rest?

At sea level a nitrogen molecule in the air has an average translational KE of \(6.2 \times 10^{-21} \mathrm{~J}\). Its mass is \(4.7 \times 10^{-26} \mathrm{~kg} .(a)\) If such a molecule could shoot straight up without striking other air molecules, how high would it rise? ( \(b\) ) What is that molecule's initial upward speed?

How much work is done on an object by the force that supports it as the object is lowered at a constant speed through a vertical distance \(h\) ? How much work does the gravitational force on the object do in this same process? The supporting force is \(m g\), where \(m\) is the mass of the object. It is directed upward while the displacement is downward. Hence, the work it does is negative: $$ F \cos \theta=(m g)(h)\left(\cos 180^{\circ}\right)=-m g h $$ The force of gravity acting on the object is also \(m g\), but it is directed downward in the same direction as the displacement. The work done on the object by the force of gravity is therefore positive: $$ F \operatorname{scos} \theta=(m g)(h)\left(\cos 0^{\circ}\right)=m g h $$

A steady force of \(500 \mathrm{~N}\) is applied horizontally to push a loaded cart at a constant speed. How far would the cart move when 3500 J of work is done on it by that applied force?

The driver of a 1200 -kg car notices that the car slows from \(20 \mathrm{~m} / \mathrm{s}\) to \(15 \mathrm{~m} / \mathrm{s}\) as it coasts a distance of \(130 \mathrm{~m}\) along level ground. How large a force opposes the motion?

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