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A \(3.2 \mathrm{~kg}\) sloth hangs \(3.0 \mathrm{~m}\) above the ground. (a) What is the gravitational potential energy of the sloth-Earth system if we take the reference point \(y=0\) to be at the ground? If the sloth drops to the ground and air drag on it is assumed to be negligible, what are the (b) kinetic energy and (c) speed of the sloth just before it reaches the ground?

Short Answer

Expert verified
Gravity potential energy is 94.08 J. Kinetic energy is also 94.08 J. Speed is about 7.67 m/s.

Step by step solution

01

Understanding Gravitational Potential Energy

Gravitational potential energy (GPE) can be calculated using the formula \( U = mgh \), where \( m \) is the mass, \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)), and \( h \) is the height above the reference point. Here, \( m = 3.2 \, \text{kg} \) and \( h = 3.0 \, \text{m} \).
02

Calculate Gravitational Potential Energy

Substitute the given values into the formula: \( U = 3.2 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 3.0 \, \text{m} \), which results in \( U = 94.08 \, \text{J} \). So, the gravitational potential energy is \( 94.08 \, \text{J} \).
03

Conversion to Kinetic Energy

When the sloth drops to the ground, all its gravitational potential energy is converted to kinetic energy because air drag is negligible. Therefore, the kinetic energy \( K \) just before reaching the ground is \( 94.08 \, \text{J} \).
04

Finding Speed from Kinetic Energy

Kinetic energy is expressed as \( K = \frac{1}{2}mv^2 \). Setting the kinetic energy equal to the gravitational potential energy: \( \frac{1}{2} \times 3.2 \, \text{kg} \times v^2 = 94.08 \, \text{J} \).
05

Solve for Speed

Solving the equation \( 1.6v^2 = 94.08 \) gives \( v^2 = \frac{94.08}{1.6} = 58.8 \). Taking the square root gives \( v = \sqrt{58.8} \approx 7.67 \, \text{m/s} \). The speed of the sloth just before reaching the ground is approximately \( 7.67 \, \text{m/s} \).

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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 an object possesses due to its motion. When we discuss kinetic energy in physics, it is often linked to how energy is transferred from one form to another, like from potential to kinetic energy. The formula to calculate it is given by \( K = \frac{1}{2}mv^2 \), where:
  • \( m \) is the mass of the object in kilograms.
  • \( v \) is the velocity of the object in meters per second.
In the original exercise, after the sloth releases its hold and drops, the gravitational potential energy is converted entirely into kinetic energy just before it reaches the ground. We calculated the kinetic energy to be \( 94.08 \, \text{J} \), which equates to the gravitational potential energy the sloth initially had. Hence, by knowing the kinetic energy, we can determine the speed of the sloth before impact using the kinetic energy formula. This demonstrates how conservation of energy enables us to understand and solve physics problems about moving objects.
Conservation of Energy
The principle of conservation of energy is paramount in physics and states that energy cannot be created or destroyed, only transformed from one form to another. In the absence of non-conservative forces, such as air resistance or friction, the total mechanical energy of a system remains constant. In the sloth problem:
  • The gravitational potential energy at the top is entirely converted into kinetic energy as it falls, assuming no loss to air resistance.
  • This conservation means the energy the sloth started with (potential) is equal to the energy it ends with just before collision (kinetic).
Understanding conservation laws allows physicists to track energy transitions in systems and is critical in scenarios ranging from simple drops to complex mechanics of moving machines. This concept helps us solve the problem with ease as it assures us that all potential energy we calculated is what becomes kinetic energy as the sloth drops.
Physics Problem Solving
Successful physics problem-solving requires the application of fundamental concepts, breaking down complex problems into manageable steps and logically analyzing each part. Let's see how these problem-solving skills were applied:
  • **Identification of Forces and Energy Types:** Start by recognizing all forms of energies and forces in the system, like gravitational force and potential energy due to height.
  • **Formulating Equations:** Use known formulas to compute quantities. Here, the potential energy and kinetic energy equations were used effectively.
  • **Calculation and Substitution:** Substituting values into equations accurately is crucial, such as placing the sloth's mass and height into the potential energy formula.
  • **Logical Analysis to Find Outcomes:** Relate these equations to solve for unknowns, like speed from kinetic energy. This involves checking your work at each step to ensure accuracy.
By understanding these steps better, students can effectively approach and solve physics problems not just limited to this exercise but many others involving energy transformations.

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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?

A \(20 \mathrm{~kg}\) block on a horizontal surface is attached to a horizontal spring of spring constant \(k=4.0 \mathrm{kN} / \mathrm{m}\). The block is pulled to the right so that the spring is stretched \(10 \mathrm{~cm}\) beyond its relaxed length, and the block is then released from rest. The frictional force between the sliding block and the surface has a magnitude of \(80 \mathrm{~N}\). (a) What is the kinetic energy of the block when it has moved \(2.0\) \(\mathrm{cm}\) from its point of release? (b) What is the kinetic energy of the block when it first slides back through the point at which the spring is relaxed? (c) What is the maximum kinetic energy attained by the block as it slides from its point of release to the point at which the spring is relaxed?

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. 8-42). (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?

A \(1.50 \mathrm{~kg}\) snowball is shot upward at an angle of \(34.0^{\circ}\) to the horizontal with an initial speed of \(20.0 \mathrm{~m} / \mathrm{s}\). (a) What is its initial kinetic energy? (b) By how much does the gravitational potential energy of the snowball-Earth system change as the snowball moves from the launch point to the point of maximum height? (c) What is that maximum height?

Snowball is fired from a cliff \(12.5 \mathrm{~m}\) high. The snowball's initial velocity is \(14.0 \mathrm{~m} / \mathrm{s}\), directed \(41.0^{\circ}\) above the horizontal. (a) How much work is done on the snowball by the gravitational force during its flight to the flat ground below the cliff? (b) What is the change in the gravitational potential energy of the snowball-Earth system during the flight? (c) If that gravitational potential energy is taken to be zero at the height of the cliff, what is its value when the snowball reaches the ground?

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