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Chapter 8: Problem 101

A \(0.63\) kg ball thrown directly upward with an initial speed of \(14 \mathrm{~m} / \mathrm{s}\) reaches a maximum height of \(8.1 \mathrm{~m}\). What is the change in the mechanical energy of the ball-Earth system during the ascent of the ball to that maximum height?

Short Answer

Expert verified
The change in mechanical energy is -11.69 J.

Step by step solution

01

Determine Initial Mechanical Energy

The initial mechanical energy of the ball includes kinetic energy only since it is at ground level (assuming). Use the kinetic energy formula: \[ KE_i = \frac{1}{2} mv^2 \]where \( m = 0.63 \space \text{kg} \) and \( v = 14 \space \text{m/s} \). Substitute these values into the formula:\[ KE_i = \frac{1}{2} \times 0.63 \times 14^2 = 61.74 \space \text{J} \]
02

Calculate Final Mechanical Energy

At the maximum height, all the initial kinetic energy is converted into potential energy. Use the gravitational potential energy formula:\[ PE_f = mgh \]where \( g = 9.81 \space \text{m/s}^2 \) and \( h = 8.1 \space \text{m} \). Substitute these values into the formula:\[ PE_f = 0.63 \times 9.81 \times 8.1 = 50.05 \space \text{J} \]
03

Find the Change in Mechanical Energy

The change in mechanical energy is the difference between the initial kinetic energy and the final potential energy:\[ \Delta E = PE_f - KE_i \]Substitute the values obtained:\[ \Delta E = 50.05 - 61.74 = -11.69 \space \text{J} \]
04

Conclusion

The negative sign indicates that there was a loss in mechanical energy, which is possibly due to non-conservative forces such as air resistance acting on the ball.

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Key Concepts

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

Kinetic Energy
Kinetic energy refers to the energy that an object possesses due to its motion. When an object moves, this type of energy quantifies the amount of work that the object can perform owing to its velocity. In the exercise, the ball possesses kinetic energy when it is thrown upwards with an initial speed.
  • The formula for kinetic energy is given by: \[ KE = \frac{1}{2} mv^2 \] where \( m \) is the mass of the object, and \( v \) is its velocity.
  • Substituting the values for the ball, with a mass of \( 0.63 \) kg and velocity \( 14 \) m/s, the initial kinetic energy calculated was \( 61.74 \) J.
Thus, at its initial state, the ball has a significant amount of kinetic energy as it moves upwards at a considerable speed.
Potential Energy
Potential energy is the energy stored in an object due to its position in a force field, like gravity. When a ball is thrown upwards, it begins to lose kinetic energy and gain potential energy as it ascends.
  • The formula for gravitational potential energy is: \[ PE = mgh \]
  • Here, \( m \) is the mass of the object, \( g \) is the acceleration due to gravity \((9.81 \, \text{m/s}^2)\), and \( h \) is the height.
  • In our problem, when the ball reaches a height of \( 8.1 \) m, its potential energy becomes \( 50.05 \) J.
This energy is fully derived from the kinetic energy initially possessed by the ball, revealing the transformation from one form of mechanical energy to another.
Energy Conservation
The principle of energy conservation states that the total energy in a closed system remains constant. In cases without external interference, potential energy and kinetic energy exchange roles but do not disappear.
  • Initially, the ball has full kinetic energy at the ground level.
  • As it reaches its maximum height, that kinetic energy becomes potential energy.
However, in realistic scenarios, like in the given problem, some energy is lost due to non-conservative forces, which causes the total mechanical energy to decrease. This observation is evident through the computed change in energy of \(-11.69\) J.
Non-Conservative Forces
Non-conservative forces are forces where energy is not conserved within the initial and final states. These include friction or air resistance that cause energy to dissipate as heat or sound.
  • The negative energy change of \(-11.69\) J in the ball's ascent demonstrates this energy loss.
  • When the ball is thrown, air resistance acts against the direction of motion, pulling energy away from the system.
Thus, even if kinetic energy is transformed completely into potential energy, non-conservative forces can reduce the total mechanical energy in the system.

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Most popular questions from this chapter

To make a pendulum, a \(300 \mathrm{~g}\) ball is attached to one end of a string that has a length of \(1.4 \mathrm{~m}\) and negligible mass. (The other end of the string is fixed.) The ball is pulled to one side until the string makes an angle of \(30.0^{\circ}\) with the vertical; then (with the string taut) the ball is released from rest. Find (a) the speed of the ball when the string makes an angle of \(20.0^{\circ}\) with the vertical and (b) the maximum speed of the ball. (c) What is the angle between the string and the vertical when the speed of the ball is one-third its maximum value?

A swimmer moves through the water at an average speed of \(0.22 \mathrm{~m} / \mathrm{s}\). The average drag force is \(110 \mathrm{~N}\). What average power is required of the swimmer?

A \(20 \mathrm{~kg}\) object is acted on by a conservative force given by \(F=-3.0 x-5.0 x^{2}\), with \(F\) in newtons and \(x\) in meters. Take the potential energy associated with the force to be zero when the object is at \(x=0 .\) (a) What is the potential energy of the system associated with the force when the object is at \(x=2.0 \mathrm{~m} ?(\mathrm{~b})\) If the object has a velocity of \(4.0 \mathrm{~m} / \mathrm{s}\) in the negative direction of the \(x\) axis when it is at \(x=5.0 \mathrm{~m}\), what is its speed when it passes through the origin? (c) What are the answers to (a) and (b) if the potential energy of the system is taken to be \(-8.0 \mathrm{~J}\) when the object is at \(x=0\) ?

A particle can move along only an \(x\) axis, where conservative forces act on it (Fig. 8-66 and the following table). The particle is released at \(x=5.00 \mathrm{~m}\) with a kinetic energy of \(K=14.0 \mathrm{~J}\) and \(\mathrm{a}\) potential energy of \(U=0\). If its motion is in the negative direction of the \(x\) axis, what are its (a) \(K\) and (b) \(U\) at \(x=2.00 \mathrm{~m}\) and its (c) \(K\) and \((\) d \() U\) at \(x=0\) ? If its motion is in the positive direction of the \(x\) axis, what are its (e) \(K\) and (f) \(U\) at \(x=11.0 \mathrm{~m}\), its \((\mathrm{g}) \mathrm{K}\) and (h) \(U\) at \(x=12.0 \mathrm{~m}\), and its (i) \(K\) and \((\mathrm{j}) U\) at \(x=13.0 \mathrm{~m} ?(\mathrm{k})\) Plot \(U(x)\) versus \(x\) for the range \(x=0\) to \(x=13.0 \mathrm{~m}\). Next, the particle is released from rest at \(x=0\). What are (l) its kinetic energy at \(x=5.0 \mathrm{~m}\) and \((\mathrm{m})\) the maximum positive position \(x_{\max }\) it reaches? (n) What does the particle do after it reaches \(x_{\max } ?\) $$ \begin{array}{lc} \hline {\text { Range }} & {\text { Force }} \\ \hline 0 \text { to } 2.00 \mathrm{~m} & \vec{F}_{1}=+(3.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 2.00 \mathrm{~m} \text { to } 3.00 \mathrm{~m} & \vec{F}_{2}=+(5.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 3.00 \mathrm{~m} \text { to } 8.00 \mathrm{~m} & F=0 \\ 8.00 \mathrm{~m} \text { to } 11.0 \mathrm{~m} & \vec{F}_{3}=-(4.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 11.0 \mathrm{~m} \text { to } 12.0 \mathrm{~m} & \vec{F}_{4}=-(1.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 12.0 \mathrm{~m} \text { to } 15.0 \mathrm{~m} & F=0 \\ \hline \end{array} $$

A \(1500 \mathrm{~kg}\) car starts from rest on a horizontal road and gains a speed of \(72 \mathrm{~km} / \mathrm{h}\) in \(30 \mathrm{~s}\). (a) What is its kinetic energy at the end of the \(30 \mathrm{~s}\) ? (b) What is the average power required of the car during the 30 s interval? (c) What is the instantaneous power at the end of the 30 s interval, assuming that the acceleration is constant?
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