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Chapter 6: Problem 14

Figure \(6-3\) shows a bead sliding on a wire. If friction forces are negligible and the bead has a speed of \(200 \mathrm{~cm} / \mathrm{s}\) at \(A\), what will be its speed \((a)\) at point \(B ?\) (b) At point \(C\) ? The energy of the bead is conserved, so we can write Change in \(\mathrm{KE}+\) change in \(\mathrm{PE}_{\mathrm{G}}=0\) $$ \frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2}+m g\left(h_{f}-h_{i}\right)=0 $$ (a) Here, \(v_{i}=2.0 \mathrm{~m} / \mathrm{s}, h_{i}=0.80 \mathrm{~m}\), and \(h_{f}=0\). Using these values, while noticing that \(m\) cancels out, gives \(v_{f}=4.4 \mathrm{~m} / \mathrm{s}\) (b) Here, \(v_{b}=2.0 \mathrm{~m} / \mathrm{s}, h_{i}=0.80 \mathrm{~m}\), and \(h_{f}=0.50 \mathrm{~m} .\) Using these values leads to \(v_{f}=3.1 \mathrm{~m} / \mathrm{s}\). Alternative Method Since energy, \(\mathrm{E}\), is conserved, $$ \begin{aligned} \mathrm{E}_{i} &=\mathrm{E}_{f} \\ \mathrm{KE}_{i}+\mathrm{PE}_{\mathrm{G}} &=\mathrm{KE}_{f}+\mathrm{PE}_{\mathrm{G} f} \\ \frac{1}{2} m v_{i}^{2}+m g h_{i} &=\frac{1}{2} m v_{f}^{2}+m g h_{f} \\ \frac{1}{2} v_{1}^{2}+g h_{i} &=\frac{1}{2} v_{f}^{2}+g h_{f} \\ (2.00 \mathrm{~m} / \mathrm{s})^{2}+2\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)\left(0.80 \mathrm{~m}-h_{f}\right)=v_{f}^{2} \end{aligned} $$

Short Answer

Expert verified
At B, speed is 4.4 m/s; at C, speed is 3.1 m/s.

Step by step solution

01

Understanding the Problem

We have a bead sliding down a wire. We're given initial conditions at point A and are required to find the speeds at point B and point C using the conservation of energy principles.
02

Applying Conservation of Energy

According to the conservation of energy, the total mechanical energy of the system remains constant if friction is negligible. Therefore, the change in kinetic energy (KE) plus the change in gravitational potential energy (PE) should equal zero.
03

Setting Up the Equation for Point B

For point B, we have initial speed \( v_i = 2.0 \text{ m/s} \) and initial height \( h_i = 0.80 \text{ m} \). The final height \( h_f \) at point B is 0 m. Apply \[ \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 + m g (h_f - h_i) = 0 \] where the mass \( m \) cancels out.
04

Solving for Final Speed at B

Since \( h_f = 0 \) at B, the equation simplifies to \( \frac{1}{2} v_f^2 = \frac{1}{2} v_i^2 + g h_i \). Solving for \( v_f \), we substitute \( v_i = 2.0 \text{ m/s} \) and \( h_i = 0.80 \text{ m} \), resulting in \( v_f = 4.4 \text{ m/s} \).
05

Setting Up the Equation for Point C

For point C, initial conditions are the same: \( v_i = 2.0 \text{ m/s} \) and \( h_i = 0.80 \text{ m} \), but now \( h_f = 0.50 \text{ m} \). Apply the same energy conservation equation.
06

Solving for Final Speed at C

The equation becomes \( \frac{1}{2} v_f^2 = \frac{1}{2} v_i^2 + g (h_i - h_f) \). Substituting \( v_i = 2.0 \text{ m/s}, h_i = 0.80 \text{ m}, \) and \( h_f = 0.50 \text{ m} \), leads to \( v_f = 3.1 \text{ m/s} \).

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Key Concepts

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

Kinetic Energy
Kinetic energy is a form of energy that an object possesses due to its motion. It is directly proportional to the mass of the object and the square of its velocity. When considering a moving object, it's important to know that the kinetic energy is given by the formula: \[ KE = \frac{1}{2} mv^2 \] where:
  • \(m\) is the mass of the object,
  • \(v\) is the velocity of the object.
At the beginning of the exercise, the bead has a velocity of \(2.0 \text{ m/s} \) at point A. This initial velocity contributes to its initial kinetic energy. As the bead moves along the wire, its velocity may change, affecting its kinetic energy as a result. The conservation of energy principle means that changes in kinetic energy must be offset by changes in potential energy, given negligible frictional forces in the scenario.
Potential Energy
Potential energy, specifically gravitational potential energy in this context, is the energy stored in an object as a result of its vertical position or height. This energy is given by:\[ PE = mgh \]where:
  • \(m\) is the mass,
  • \(g\) is the acceleration due to gravity (\(9.81 \text{ m/s}^2 \)),
  • \(h\) is the height of the object above a reference point.
In the exercise, the bead is at different heights at different points along the wire. Initially, at point A, the bead has an initial height \(h_i = 0.80 \text{ m}\). As it moves to points B and C, the heights change, which affects the potential energy. At point B, the height is \(0 \text{ m}\), meaning all potential energy is converted to kinetic energy. At point C, the bead has some height \(0.50 \text{ m}\), retaining some gravitational potential energy.
Mechanical Energy
Mechanical energy is the sum of kinetic and potential energies in the system. For systems where only conservative forces (like gravity) are acting, mechanical energy is conserved. This is illustrated by the equation:\[ E = KE + PE \]In our problem, since friction is negligible, the mechanical energy at any two points along the bead鈥檚 path is constant. This means:\[ KE_{i} + PE_{i} = KE_{f} + PE_{f} \]Here, \(i\) refers to initial conditions (at point A, for example), and \(f\) refers to final conditions (at either B or C). This conservation makes it possible to find unknown variables, like velocity, when given the system鈥檚 initial mechanical energy.
Problem Solving Steps
Identifying the logical steps in solving this type of energy conservation problem is crucial:
  • Step 1: Begin by recognizing that energy conservation applies. This means the total initial energy equals the total final energy.
  • Step 2: Use the known initial conditions to calculate the initial mechanical energy: both kinetic and potential.
  • Step 3: Set up the equation \(\frac{1}{2} mv_{f}^2 - \frac{1}{2} mv_{i}^2 + mg(h_{f} - h_{i}) = 0\), cancel out the mass \(m\), if possible, for simplicity.
  • Step 4: Substitute the known values and solve for the unknown (usually final velocity \(v_f\)).
In this problem, these steps enable you to determine the bead's velocities at different points, B and C, by applying the conservation of mechanical energy concept.

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Most popular questions from this chapter

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.

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 \(m g h\).

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 \(s=274 \mathrm{~m}=0.27 \mathrm{~km}\).

In Fig. \(6-1\), assume that the object is being pulled in a straight line along the ground by a \(75-\mathrm{N}\) force directed \(28^{\circ}\) above the horizontal. How much work does the force do in pulling the object \(8.0 \mathrm{~m}\) ? The work done is equal to the product of the displacement, \(8.0 \mathrm{~m}\), and the component of the force that is parallel to the displacement, \((75 \mathrm{~N})\left(\cos 28^{\circ}\right)\). Thus, $$ W=(75 \mathrm{~N})\left(\cos 28^{\circ}\right)(8.0 \mathrm{~m})=0.53 \mathrm{~kJ} $$
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