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A \(0.63 \mathrm{~kg}\) 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

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Step by step solution

01

Understanding the Problem

We need to find the change in mechanical energy of the ball-Earth system from the moment the ball is thrown till it reaches its maximum height.
02

Calculate Initial Mechanical Energy

The initial mechanical energy is the sum of kinetic energy and potential energy at the starting point. The initial potential energy is 0 because it is thrown from the ground. The initial kinetic energy is given by the formula: \( KE = \frac{1}{2}mv^2 \). Where \( m = 0.63 \, \text{kg} \) and \( v = 14 \, \text{m/s} \).

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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 the energy that an object possesses due to its motion. Picture a ball being thrown upward; as it moves, it carries kinetic energy. This can be calculated with the formula: \[ KE = \frac{1}{2}mv^2 \]Here, \( m \) represents the mass of the object, and \( v \) represents its velocity. In our case, the ball has a mass of \( 0.63 \, \text{kg} \) and an initial speed of \( 14 \, \text{m/s} \).
  • Kinetic energy is significant when an object is in motion.
  • The faster the object moves, the higher its kinetic energy.
  • For the ball, its kinetic energy when thrown is \[ KE_{\text{initial}} = \frac{1}{2} \times 0.63 \, \text{kg} \times (14 \, \text{m/s})^2 \]
As the ball ascends and slows down, its kinetic energy decreases, converting to potential energy.
Potential Energy
Potential energy is the stored energy of an object based on its position or height. When the ball reaches its maximum height, its kinetic energy becomes negligible and it has maximum potential energy. The potential energy has increased as the ball moved against the gravitational force.
  • Potential energy depends on the height above the ground and gravitational acceleration.
  • The formula to compute gravitational potential energy is \[ PE = mgh \] where \( m \) is mass, \( g \) is the gravitational acceleration (approximately \( 9.8 \, \text{m/s}^2 \) on Earth), and \( h \) is the height.
In this scenario, when the ball is at its maximum height of \( 8.1 \, \text{m} \), its potential energy will be:\[ PE_{\text{final}} = 0.63 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 8.1 \, \text{m} \]This represents the energy stored due to the ball's raised position.
Conservation of Energy
The principle of conservation of energy states that the total energy in an isolated system remains constant. As the ball is thrown upward, the system transitions energy between kinetic and potential forms but the total mechanical energy is conserved.
  • Initially, the ball has high kinetic energy and low potential energy since it’s just starting its ascent from the ground.
  • As the ball reaches the top of its path, most kinetic energy has been converted into potential energy.
Mechanical energy is the sum of kinetic and potential energy. Even though they shift between each other, their total remains unchanged during the motion when air resistance and other external forces are negligible:\[ E_{\text{mechanical}} = KE + PE \]In this way, there is no net change in the mechanical energy of the ball-Earth system during its ascent, assuming no external work is done. During the ascent to its maximum height, the kinetic energy initially decreases, converting into potential energy, but the total sum remains the same.

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Most popular questions from this chapter

A \(0.50 \mathrm{~kg}\) banana is thrown directly upward with an initial speed of \(4.00 \mathrm{~m} / \mathrm{s}\) and reaches a maximum height of \(0.80 \mathrm{~m} .\) What change does air drag cause in the mechanical energy of the bananaEarth system during the ascent?

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 \(x_{\max } ?\) $$ \begin{array}{lc} {\text { Range }} & \text { Force } \\ \hline 0 \text { to } 2.00 \mathrm{~m} & \vec{F}_{1}=+(3.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 2.00 \mathrm{~m} \text { to } 3.00 \mathrm{~m} & \vec{F}_{2}=+(5.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 3.00 \mathrm{~m} \text { to } 8.00 \mathrm{~m} & F=0 \\ 8.00 \mathrm{~m} \text { to } 11.0 \mathrm{~m} & \vec{F}_{3}=-(4.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 11.0 \mathrm{~m} \text { to } 12.0 \mathrm{~m} & \vec{F}_{4}=-(1.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 12.0 \mathrm{~m} \text { to } 15.0 \mathrm{~m} & F=0 \end{array} $$ For the arrangement of forces in Problem \(81,\) a \(2.00 \mathrm{~kg}\) particle is released at \(x=5.00 \mathrm{~m}\) with an initial velocity of \(3.45 \mathrm{~m} / \mathrm{s}\) in the negative direction of the \(x\) axis. (a) If the particle can reach \(x=0 \mathrm{~m},\) what is its speed there, and if it cannot, what is its turning point? Suppose, instead, the particle is headed in the positive \(x\) direction when it is released at \(x=5.00 \mathrm{~m}\) at speed \(3.45 \mathrm{~m} / \mathrm{s} .\) (b) If the particle can reach \(x=13.0 \mathrm{~m},\) what is its speed there, and if it cannot, what is its turning point?

A block with mass \(m=2.00 \mathrm{~kg}\) is placed against a spring on a frictionless incline with angle \(\theta=30.0^{\circ}\) (Fig. 8-44). (The block is not attached to the spring.) The spring, with spring constant \(k=19.6 \mathrm{~N} / \mathrm{cm},\) is compressed \(20.0 \mathrm{~cm}\) and then released. (a) What is the elastic potential energy of the compressed spring? (b) What is the change in the gravitational potential energy of the block- Earth system as the block moves from the release point to its highest point on the incline? (c) How far along the incline is the highest point from the release point?

131 Fasten one end of a vertical spring to a ceiling, attach a cabbage to the other end, and then slowly lower the cabbage until the upward force on it from the spring balances the gravitational force on it. Show that the loss of gravitational potential energy of the cabbage-Earth system equals twice the gain in the spring's potential energy.

A \(60.0 \mathrm{~kg}\) circus performer slides \(4.00 \mathrm{~m}\) down a pole to the circus floor, starting from rest. What is the kinetic energy of the performer as she reaches the floor if the frictional force on her from the pole (a) is negligible (she will be hurt) and (b) has a magnitude of \(500 \mathrm{~N}\) ?

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