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Chapter 8: Problem 66

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
(a) 94.08 J; (b) 94.08 J; (c) approximately 7.67 m/s.

Step by step solution

01

Calculate Gravitational Potential Energy

The gravitational potential energy (GPE) is given by the formula: \( U = mgh \), where \( m \) is the mass (3.2 kg), \( g \) is the acceleration due to gravity (9.8 m/s虏), and \( h \) is the height (3.0 m). Calculate \( U \): \[ U = 3.2 \times 9.8 \times 3.0 = 94.08 \, \text{Joules} \] Thus, the gravitational potential energy of the system is \( 94.08 \, \text{J} \).
02

Determine Kinetic Energy at the Ground

When the sloth reaches the ground, all the gravitational potential energy will be converted to kinetic energy (KE) since air drag is negligible. Thus, the kinetic energy at the ground is equal to the initial gravitational potential energy:\[ \text{KE} = 94.08 \, \text{J} \]
03

Calculate Speed Just Before Hitting the Ground

Use the formula for kinetic energy: \( KE = \frac{1}{2}mv^2 \), where \( m \) is the sloth's mass (3.2 kg) and \( v \) is its speed. Set \( KE = 94.08 \, \text{J} \) and solve for \( v \):\[94.08 = \frac{1}{2} \times 3.2 \times v^2 \94.08 = 1.6v^2 \v^2 = \frac{94.08}{1.6} \v^2 = 58.8 \v = \sqrt{58.8} \v \approx 7.67 \, \text{m/s} \]Thus, 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 a type of energy that an object possesses due to its motion. When the sloth in our exercise falls from a height, it gains kinetic energy as it speeds up. Kinetic energy is dependent on two main factors:
  • Mass of the Object: The heavier the object, the more kinetic energy it can have.
  • Speed of the Object: Even a small increase in speed can significantly boost the kinetic energy.
The formula to calculate kinetic energy is given by:\[ \text{KE} = \frac{1}{2}mv^2 \]Where \(m\) is the mass and \(v\) represents the velocity of the object. During the sloth's fall, every bit of gravitational potential energy transforms into kinetic energy as there is no air resistance working against it. By the time the sloth reaches the ground, all its potential energy is converted into kinetic energy, allowing you to calculate its velocity just before impact.
Conversion of Energy
The principle of the conservation of energy states that energy in a closed system cannot be created or destroyed. It can only change forms. In our scenario, as the sloth drops, we witness a perfect example of energy conversion:When the sloth is hanging, it's loaded with gravitational potential energy due to its height above the ground. This energy is calculated using \( U = mgh \), where \( m \) is the mass, \( g \) is gravitational acceleration, and \( h \) is the height.As the sloth descends, its height鈥攈ence its potential energy鈥攄ecreases. However, the total energy remains constant. The lost potential energy converts into kinetic energy as it accelerates toward the ground. By the time it reaches the ground, all potential energy has metamorphosed into kinetic energy. Knowing this, one can equate potential energy at the top to kinetic energy at the bottom and solve for variables like speed.
Speed Calculation
The journey from potential to kinetic energy involves finding out how fast the object, like our sloth, is moving just before it hits the ground. To achieve this, you use the relationship between kinetic energy and speed. Once you know the kinetic energy, use the following steps to calculate speed:1. **Identify Known Values**: You have kinetic energy \(KE = 94.08 \, \text{J}\) and mass \(m = 3.2 \, \text{kg}\).2. **Apply the Kinetic Energy Formula**: Rearrange \( \text{KE} = \frac{1}{2}mv^2 \) to solve for \(v\).3. **Solve for Speed**: Do the math as follows: \[ 94.08 = \frac{1}{2} \times 3.2 \times v^2 \] \[ 1.6v^2 = 94.08 \] \[ v^2 = \frac{94.08}{1.6} = 58.8 \] \[ v = \sqrt{58.8} \approx 7.67 \, \text{m/s} \]The steps above show that the speed of the sloth as it reaches the ground is around \(7.67 \, \text{m/s}\), rounding out the full conversion from potential to kinetic energy during the fall.

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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 banana-Earth system during the ascent?

The potential energy of a diatomic molecule (a two-atom system like \(\mathrm{H}_{2}\) or \(\mathrm{O}_{2}\) ) is given by $$ U=\frac{A}{r^{12}}-\frac{B}{r^{6}} $$ where \(r\) is the separation of the two atoms of the molecule and \(A\) and \(B\) are positive constants. This potential energy is associated with the force that binds the two atoms together. (a) Find the equilibrium separation - that is, the distance between the atoms at which the force on each atom is zero. Is the force repulsive (the atoms are pushed apart) or attractive (they are pulled together) if their separation is (b) smaller and (c) larger than the equilibrium separation?

A certain spring is found not to conform to Hooke's law. The force (in newtons) it exerts when stretched a distance \(x\) (in meters) is found to have magnitude \(52.8 x+38.4 x^{2}\) in the direction opposing the stretch. (a) Compute the work required to stretch the spring from \(x=0.500 \mathrm{~m}\) to \(x=1.00 \mathrm{~m} .(\mathrm{b})\) With one end of the spring fixed, a particle of mass \(2.17 \mathrm{~kg}\) is attached to the other end of the spring when it is stretched by an amount \(x=1.00 \mathrm{~m}\). If the particle is then released from rest, what is its speed at the instant the stretch in the spring is \(x=0.500 \mathrm{~m} ?(\mathrm{c})\) Is the force exerted by the spring conservative or nonconservative? Explain.

A horizontal force of magnitude \(35.0 \mathrm{~N}\) pushes a block of mass \(4.00 \mathrm{~kg}\) across a floor where the coefficient of kinetic friction is \(0.600 .\) (a) How much work is done by that applied force on the block- floor system when the block slides through a displacement of \(3.00 \mathrm{~m}\) across the floor? (b) During that displacement, the thermal energy of the block increases by \(40.0 \mathrm{~J}\). What is the increase in thermal energy of the floor? (c) What is the increase in the kinetic energy of the block?

A rope is used to pull a \(3.57 \mathrm{~kg}\) block at constant speed \(4.06 \mathrm{~m}\) along a horizontal floor. The force on the block from the rope is \(7.68 \mathrm{~N}\) and directed \(15.0^{\circ}\) above the horizontal. What are (a) the work done by the rope's force, (b) the increase in thermal energy of the block-floor system, and (c) the coefficient of kinetic friction between the block and floor?
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