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Chapter 8: Problem 25

At \(t=0\) a \(1.0 \mathrm{~kg}\) ball is thrown from a tall tower with \(\vec{v}=(18 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(24 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}} .\) What is \(\Delta U\) of the ball - Earth system between \(t=0\) and \(t=6.0 \mathrm{~s}\) (still free fall)?

Short Answer

Expert verified
The change in potential energy is -317.52 J.

Step by step solution

01

Define Gravitational Potential Energy

Gravitational potential energy is the energy an object possesses due to its position in a gravitational field. It is given by the formula \( U = mgh \), where \( m \) is the mass of the object, \( g \) is the acceleration due to gravity, and \( h \) is the height above a reference point.
02

Determine Initial and Final Vertical Velocity Components

The initial vertical velocity component \( v_{0_y} \) is 24 m/s (given in the problem). To find the vertical velocity after 6 seconds (\( v_{y} \)), use the equation \( v_{y} = v_{0_y} - gt \), where \( g \) is 9.8 m/s虏 (acceleration due to gravity) and \( t = 6 \) s.
03

Solve for Final Vertical Velocity

Substitute the given values into the equation: \[ v_{y} = 24 \text{ m/s} - (9.8 \text{ m/s}^2)(6 \text{ s}) \]Calculating this gives \[ v_{y} = 24 - 58.8 = -34.8 \text{ m/s} \].
04

Calculate Initial and Final Heights Using Kinematics

Use the kinematic equation for height: \[ h = v_{0_y}t - \frac{1}{2}gt^2 \].Start by calculating the initial height \( h_0 \) (at \( t=0 \)), where height is zero:\[ h_0 = 24 \times 0 - \frac{1}{2} \times 9.8 \times 0^2 = 0 \text{ m} \].Now, calculate the height after 6 s:\[ h_6 = 24 \times 6 - \frac{1}{2} \times 9.8 \times 6^2 = 144 - 176.4 = -32.4 \text{ m} \].
05

Calculate Change in Potential Energy

The change in potential energy \( \Delta U \) is given by:\[ \Delta U = mgh_6 - mgh_0 \].Substituting the known values (\( m = 1 \text{ kg} \), \( g = 9.8 \text{ m/s}^2 \), \( h_0 = 0 \), \( h_6 = -32.4 \)), we have:\[ \Delta U = (1)(9.8)(-32.4) - (1)(9.8)(0) \]\[ \Delta U = -317.52 \text{ J} \].
06

Interpret the Result

The negative sign indicates that the potential energy of the ball-Earth system has decreased by 317.52 J as the ball falls.

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Key Concepts

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

Kinematics
Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause the motion. It primarily involves equations that describe the trajectories and velocities of moving objects.
In our context, kinematics is used to understand the motion of the ball thrown from the tower. When analyzing such motion, we focus on various kinematic parameters like velocity, acceleration, displacement, and time.
  • The horizontal motion is uniform because no force acts along the horizontal axis.
  • The vertical motion is influenced by gravity, affecting how fast the ball rises or falls with time.
Understanding kinematics is critical as it serves as the foundation for calculating other factors like energy and impact forces when an object is in motion.
Free Fall
Free fall refers to the motion of an object under the influence of gravitational force alone. This occurs when an object is dropped, thrown, or released from a height, allowing gravity to pull it downward without any air resistance.
In the given problem, the ball is considered to be in free fall conditions after being projected. This means that its vertical motion is governed solely by gravity.
During free fall:
  • The object accelerates at a constant rate due to gravity, which is approximately 9.8 m/s虏.
  • Horizontal motion does not affect the free fall as gravity acts vertically.
Understanding free fall helps determine velocity, height, and displacement as time progresses.
Vertical Velocity
The vertical velocity of an object in motion is a crucial aspect of projectile motion. It tells us how fast the object moves up or down.
In our scenario, we begin with an initial vertical velocity known as the initial vertical component, denoted as \( v_{0_y} = 24 \text{ m/s} \). As time passes, gravity continually influences this velocity, causing it to change.
  • Vertical velocity is calculated using: \( v_y = v_{0_y} - gt \), where \( g \) is the acceleration due to gravity.
  • After 6 seconds, this velocity becomes \( v_y = -34.8 \text{ m/s} \), indicating a downward direction.
Understanding how vertical velocity changes helps in calculating the trajectory of the projectile, as well as its eventual impact speed.
Energy Conservation
Energy conservation in the context of gravitational systems analyzes how energy shifts between potential and kinetic forms, staying constant overall. Gravitational potential energy and kinetic energy interplay as objects move within a gravitational field.
For the ball in free fall:
  • Initially, the ball possesses both kinetic energy and gravitational potential energy.
  • As the ball descends, potential energy decreases and is converted into kinetic energy.
  • The change in potential energy, \( \Delta U = -317.52 \text{ J} \), illustrates this conversion, showing a reduction as the ball falls.
Energy conservation underlines the principles that govern motion, even as energy types shift, maintaining equilibrium in a closed system without external forces.

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Most popular questions from this chapter

A large fake cookie sliding on a horizontal surface is attached to one end of a horizontal spring with spring constant \(k=400 \mathrm{~N} / \mathrm{m} ;\) the other end of the spring is fixed in place. The cookie has a kinetic energy of \(20.0 \mathrm{~J}\) as it passes through the spring's equilibrium position. As the cookie slides, a frictional force of magnitude \(10.0 \mathrm{~N}\) acts on it. (a) How far will the cookie slide from the equilibrium position before coming momentarily to rest? (b) What will be the kinetic energy of the cookie as it slides back through the equilibrium position?

When a click beetle is upside down on its back, it jumps upward by suddenly arching its back, transferring energy stored in a muscle to mechanical energy. This launching mechanism produces an audible click, giving the beetle its name. Videotape of a certain clickbeetle jump shows that a beetle of mass \(m=4.0 \times 10^{-6} \mathrm{~kg}\) moved directly upward by \(0.77 \mathrm{~mm}\) during the launch and then to a maximum height of \(h=0.30 \mathrm{~m}\). During the launch, what are the average magnitudes of (a) the external force on the beetle's back from the floor and (b) the acceleration of the beetle in terms of \(g ?\)

Figure 8 -31 shows a ball with mass \(m=0.341 \mathrm{~kg}\) attached to the end of a thin rod with length \(L=0.452 \mathrm{~m}\) and negligible mass. The other end of the rod is pivoted so that the ball can move in a vertical circle. The rod is held horizontally as shown and then given enough of a downward push to cause the ball to swing down and around and just reach the vertically up position, with zero speed there. How much work is done on the ball by the gravitational force from the initial point to (a) the lowest point, (b) the highest point, and (c) the point on the right level with the initial point? If the gravitational potential energy of the ball-Earth system is taken to be zero at the initial point, what is it when the ball reaches (d) the lowest point, (e) the highest point, and (f) the point on the right level with the initial point? (g) Suppose the rod were pushed harder so that the ball passed through the highest point with a nonzero speed. Would \(\Delta U_{g}\) from the lowest point to the highest point then be greater than, less than, or the same as it was when the ball stopped at the highest point?

The summit of Mount Everest is \(8850 \mathrm{~m}\) above sea level. (a) How much energy would a \(90 \mathrm{~kg}\) climber expend against the gravitational force on him in climbing to the summit from sea level? (b) How many candy bars, at 1.25 MJ per bar, would supply an energy equivalent to this? Your answer should suggest that work done against the gravitational force is a very small part of the energy expended in climbing a mountain.

A spring \((k=200 \mathrm{~N} / \mathrm{m})\) is fixed at the top of a frictionless plane inclined at angle \(\theta=40^{\circ}(\) Fig. \(8-59) .\) A \(1.0 \mathrm{~kg}\) block is projected up the plane, from an initial position that is distance \(d=0.60 \mathrm{~m}\) from the end of the relaxed spring, with an initial kinetic energy of 16 J. (a) What is the kinetic energy of the block at the instant it has compressed the spring \(0.20 \mathrm{~m} ?\) (b) With what kinetic energy must the block be projected up the plane if it is to stop momentarily when it has compressed the spring by \(0.40 \mathrm{~m} ?\)
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