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Chapter 6: Problem 3

A \(300-\mathrm{g}\) object slides \(80 \mathrm{~cm}\) 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 \(F_{\mathrm{t}} s \cos \theta\). Because the friction force is opposite in direction to the displacement, \(\theta=180^{\circ} .\) Therefore, $$ \text { Work }=F_{f} s \cos 180^{\circ}=(0.588 \mathrm{~N})(0.80 \mathrm{~m})(-1)=-0.47 \mathrm{~J} $$ 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 in overcoming friction is \(-0.47 \text{ J}.\)

Step by step solution

01

Calculate Normal Force

The first step in finding the work done by friction is to calculate the normal force. Since the object is moving horizontally and there is no vertical motion, the normal force (\( F_N \)) equals the weight of the object. The weight is calculated using the formula \( F_N = m \cdot g \), where \( m = 0.300 \text{ kg} \) (the mass of the object) and \( g = 9.81 \text{ m/s}^2 \) (acceleration due to gravity). Thus, \( F_N = (0.300 \text{ kg})(9.81 \text{ m/s}^2) = 2.943 \text{ N}. \)
02

Calculate Friction Force

The friction force \( F_f \) can be calculated using the formula \( F_f = \mu_k \cdot F_N \), where \( \mu_k = 0.20 \) is the coefficient of kinetic friction. Therefore, \( F_f = (0.20)(2.943 \text{ N}) = 0.588 \text{ N}. \)
03

Convert Distance to Meters

The problem states that the object slides 80 cm. Since the standard unit of distance in this context is meters, we convert it to meters: \( 80 \text{ cm} = 0.80 \text{ m}. \)
04

Calculate Work Done by Friction

The work done by friction is calculated using the formula \( \text{Work} = F_f \cdot s \cdot \cos \theta \), where \( \theta \) is the angle between the force and the direction of motion. Here, \( \theta = 180^{\circ} \), because the frictional force opposes the motion. Thus, \( \cos 180^{\circ} = -1 \). So \( \text{Work} = (0.588 \text{ N})(0.80 \text{ m})(-1) = -0.47 \text{ J}. \) The negative sign indicates that the work is done against the direction of motion.

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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 the force that opposes the motion of two surfaces sliding past each other. In this problem, we are dealing with an object moving along a tabletop where kinetic friction acts to slow it down. The kinetic friction force can be determined using:
  • The coefficient of kinetic friction (\(\mu_{k}\)) which is a constant specific to the pair of materials in contact.
  • The normal force (\(F_N\)), which is the force exerted perpendicular to the surfaces in contact.
The formula to calculate kinetic friction is:\[ F_f = \mu_k \cdot F_N \] Thus, in the example provided, the kinetic friction force was calculated as 0.588 N, opposing the motion of the object and acting along the plane of the tabletop. Overcoming this friction means doing work against this force, which is what we are trying to quantify.
Normal Force
Normal force is the perpendicular contact force exerted by a surface on an object resting on it. For an object moving horizontally, the normal force is equal to the weight of the object provided that there are no other vertical forces at play. It is crucial in calculating kinetic friction since it determines how strongly two surfaces are pressed together.
  • Weight is calculated using mass and gravity: \[ F_N = m \times g \]
  • In our scenario, it is calculated as: \[ F_N = 0.300 \text{ kg} \times 9.81 \text{ m/s}^2 = 2.943 \text{ N} \]
This step is critical because it ensures that kinetic friction is calculated accurately since it directly influences the frictional force exerted by the surface on the object.
Work-Energy Principle
The work-energy principle connects the work done on an object to its energy. Specifically, it states that the work done by forces results in a change in kinetic energy. In the context of kinetic friction, this principle helps us relate the work done by friction to changes in the object's motion.
  • The formula used is: \[ \text{Work} = F_f \cdot s \cdot \cos \theta \]
  • Here, the work done is negative, equalling \(-0.47 \text{ J}\), due to the friction force opposing motion.
The negative value indicates an energy loss from the system as friction converts kinetic energy into other forms, often heat, effectively resisting motion and slowing down the object.
Angle of Displacement
The angle of displacement refers to the angle between the direction of force applied and the direction of the object's movement. This angle affects the calculation of work done by friction.
  • In physics problems like ours, often look for whether forces and movements are aligned or opposed.
  • When the angle is \(180^{\circ}\), force and displacement vectors point in exactly opposite directions, leading to: \[ \cos 180^{\circ} = -1 \]
This results in negative work as seen here. Understanding this angle is crucial for determining how forces interact with an object's motion, thereby impacting energy calculations significantly and correctly characterizing the nature of the motion.

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Most popular questions from this chapter

Calculate the average power required to raise a \(150-\mathrm{kg}\) drum to a height of \(20 \mathrm{~m}\) in a time of \(1.0\) minute. Give your answer in both kilowatts and horsepower.

In an Atwood machine (see Problem \(3.30\) ), the two masses are \(800 \mathrm{~g}\) and \(700 \mathrm{~g}\). The system is released from rest. How fast is the \(800-\mathrm{g}\) mass moving after it has fallen \(120 \mathrm{~cm}\) ? The \(700-\mathrm{g}\) mass rises \(120 \mathrm{~cm}\) while the \(800-\mathrm{g}\) mass falls \(120 \mathrm{~cm}\), so the net change in \(\mathrm{PE}_{\mathrm{G}}\) is $$ \text { Change in } \mathrm{PE}_{\mathrm{G}}=(0.70 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(1.20 \mathrm{~m})-(0.80 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(1.20 \mathrm{~m})=-1.18 \mathrm{~J} $$ which is a loss in \(\mathrm{PE}_{\mathrm{G}}\). Because energy is conserved, the \(\mathrm{KE}\) of the masses must increase by \(1.18 \mathrm{~J}\). Therefore, $$ \text { Change in } \mathrm{KE}=1.18 \mathrm{~J}=\frac{1}{2}(0.70 \mathrm{~kg})\left(v_{f}^{2}-v_{i}^{2}\right)+\frac{1}{2}(0.80 \mathrm{~kg})\left(v_{f}^{2}-v_{i}^{2}\right) $$ The system started from rest, so \(v_{i}=0\). We solve the above equation for \(v_{f}\) and find \(v_{f}=1.25 \mathrm{~m} / \mathrm{s}\).

How much work is done against gravity in lifting a \(3.0-\mathrm{kg}\) object through a vertical distance of \(40 \mathrm{~cm}\) ? An external force is needed to lift an object. If the object is raised at constant speed, the lifting force must equal the weight of the object. The work done by the lifting force is referred to as work done against gravity. Because the lifting force is \(m g\), where \(m\) is the mass of the object, $$ \text { Work }=(m g)(h)(\cos \theta)=(3.0 \mathrm{~kg} \times 9.81 \mathrm{~N})(0.40 \mathrm{~m})(1)=12 \mathrm{~J} $$ In general, the work done against gravity in lifting an object of mass \(m\) through a vertical distance \(h\) is \(m g h\).

An advertisement claims that a certain \(1200-\mathrm{kg}\) car can accelerate from rest to a speed of \(25 \mathrm{~m} / \mathrm{s}\) in a time of \(8.0 \mathrm{~s}\). What average power must the motor develop to produce this acceleration? Give your answer in both watts and horsepower. Ignore friction losses. The work done in accelerating the car is $$ \text { Work done }=\text { Change in } \mathrm{KE}=\frac{1}{2} m\left(v_{f}^{2}-v_{i}^{2}\right)=\frac{1}{2} m v_{f}^{2} $$ The time taken for this work to be performed is \(8.0 \mathrm{~s}\). Therefore, to two significant figures, $$ \text { Power }=\frac{\text { Work }}{\text { Time }}=\frac{\frac{1}{2}(1200 \mathrm{~kg})(25 \mathrm{~m} / \mathrm{s})^{2}}{8.0 \mathrm{~s}}=46875 \mathrm{~W}=47 \mathrm{~kW} $$ Converting from watts to horsepower, we have $$ \text { Power }=(46875 \mathrm{~W})\left(\frac{1 \mathrm{hp}}{746 \mathrm{~W}}\right)=63 \mathrm{hp} $$

A \(60000-\mathrm{kg}\) train is being dragged along a straight line up a \(1.0\) percent grade (i.e., the road rises \(1.0 \mathrm{~m}\) for each \(100 \mathrm{~m}\) traveled horizontally) by a steady drawbar pull of \(3.0 \mathrm{kN}\) parallel to the incline. The friction force opposing the motion of the train is \(4.0 \mathrm{kN}\). The train's initial speed is \(12 \mathrm{~m} / \mathrm{s}\). Through what distance \(s\) will the train move along its tracks before its speed is reduced to \(9.0 \mathrm{~m} / \mathrm{s}\) ? Use energy considerations. The change in total energy of the train is due to the work done by the friction force (which is negative) and the drawbar pull (which is positive): Change in \(\mathrm{KE}+\) change in \(\mathrm{PE}_{\mathrm{G}}=W_{\text {drombur }}+W_{\text {triction }}\) The train loses \(\mathrm{KE}\) and gains \(\mathrm{PE}_{\mathrm{G}}\). It rises a height \(h=s \sin \theta\), where \(\theta\) is the incline angle and \(\tan \theta=1 / 100\). Hence, \(\theta=0.573^{\circ}\), and \(h=0.010 s\) (at small angles \(\left.\tan \theta \approx \sin \theta\right)\). Therefore, $$ \begin{array}{r} \frac{1}{2} m\left(v_{f}^{2}-v_{i}^{2}\right)+m g(0.010 s)=(3000 \mathrm{~N})(s)(1)+(4000 \mathrm{~N})(s)(-1) \\ -1.89 \times 10^{6} \mathrm{~J}+\left(5.89 \times 10^{3} \mathrm{~N}\right) s=(-1000 \mathrm{~N}) s \end{array} $$ from which we obtain \(s=274 \mathrm{~m}=0.27 \mathrm{~km}\).
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