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Chapter 6: Problem 18

A 4.80 -kg watermelon is dropped from rest from the roof of a 25.0 -m-tall building and feels no appreciable air resistance. (a) Calculate the work done by gravity on the watermelon during its displacement from the roof to the ground. (b) Just before it strikes the ground, what is the watermelon's (i) kinetic energy and (ii) speed? (c) Which of the answers in parts (a) and (b) would be different if there were appreciable air resistance?

Short Answer

Expert verified
(a) Work done by gravity is 1177.8 J. (b) (i) KE is 1177.8 J, (ii) speed is 22.0 m/s. (c) KE and speed change with air resistance.

Step by step solution

01

Calculate the work done by gravity (a)

The work done by gravity can be calculated using the formula \( W = mgh \), where \( m \) is the mass, \( g \) is the acceleration due to gravity (approximately \( 9.81 \text{ m/s}^2 \)), and \( h \) is the height. Plugging in the values, we get: \( W = 4.80 \times 9.81 \times 25.0 \). Calculating this gives the work done by gravity as \( W = 1177.8 \text{ J} \).
02

Calculate kinetic energy before impact (b)(i)

Just before it strikes the ground, the kinetic energy (K.E) of the watermelon is equal to the work done by gravity because it started from rest and feels no air resistance. Thus, \( K.E = 1177.8 \text{ J} \).
03

Calculate the speed just before impact (b)(ii)

Using the kinetic energy formula \( K.E = \frac{1}{2} mv^2 \), where \( m \) is the mass and \( v \) is the velocity. Rearrange to solve for \( v \): \( v = \sqrt{\frac{2K.E}{m}} \). Substitute \( K.E = 1177.8 \text{ J} \) and \( m = 4.80 \text{ kg} \). Calculating gives the speed \( v = \sqrt{\frac{2 \times 1177.8}{4.80}} = 22.0 \text{ m/s} \).
04

Determine effects of air resistance (c)

If there were appreciable air resistance, the work done by gravity (a) would remain the same because it only depends on the height, mass, and gravitational acceleration. However, both the kinetic energy (b)(i) and speed (b)(ii) just before impact would be less due to energy loss from air resistance.

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Key Concepts

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

Kinetic Energy: Understanding the Basics
Kinetic energy (K.E) is the energy that an object possesses due to its motion. Imagine riding a bicycle down a hill. When you're moving faster at the bottom than at the top, you have more kinetic energy at the bottom. The formula for kinetic energy is given by: \( K.E = \frac{1}{2} mv^2 \), where \( m \) represents mass, and \( v \) is velocity.
In the case of the watermelon, just before striking the ground, it has its maximum kinetic energy. This is because it was dropped from rest, converting all potential energy due to height into kinetic energy, assuming no air resistance. This conversion makes kinetic energy equal to the work done by gravity in this scenario.
  • Formula: \( K.E = \frac{1}{2} mv^2 \)
  • The higher the velocity, the greater the kinetic energy.
  • In the exercise, \( K.E = 1177.8 \text{ J} \) as all potential energy converted to kinetic energy.
Gravity: The Constant Force
Gravity is the force that pulls objects toward the center of the Earth or any other celestial body. It's what keeps us grounded and dictates how objects fall when dropped. This force is consistent, known as the gravitational acceleration \( g \) which is approximately \( 9.81 \text{ m/s}^2 \) on Earth.
In physics problems, gravity aids in calculating the work done on an object in free fall. In the watermelon problem, gravity does work on the dropped object, causing it to accelerate downward, increasing its speed and kinetic energy just before hitting the ground. Work done by gravity is given by \( W = mgh \), where \( h \) is the height.
  • Gravity's role is integral in determining how fast an object speeds up when falling.
  • \( W = 4.80 \times 9.81 \times 25 = 1177.8 \text{ J} \)
  • This work done results in the same amount of kinetic energy at the lowest point, assuming no air resistance.
Air Resistance: Nature's Brake
Air resistance is a force that opposes the motion of objects through the air. Imagine trying to run fast with your hands spread wide. You'll feel the air pushing against you, making it harder to speed up. It's the same for objects falling through the air.
While the exercise specifies no appreciable air resistance, it's important to understand its effects. An object experiencing air resistance loses some energy to the air, meaning not all gravitational work converts to kinetic energy. This would result in lower speed and kinetic energy just before impact.
  • Reduces speed and kinetic energy of falling objects.
  • Especially strong in lighter or larger surface area objects.
  • In real-world scenarios, consider air resistance for more accurate calculations.
Physics Problem Solving: Effective Strategies
Solving physics problems requires a systematic approach. Breaking problems into steps helps make complex ideas manageable. Begin by understanding the problem, identifying forces, and recognizing key variables and equations to use.
For example, with the watermelon, we identify the forces at play and the relevant formulas: gravitational work and kinetic energy calculations. It's crucial to follow a step-by-step method to ensure no part of the problem is overlooked. Similarly, anticipate influences like air resistance, even if initially neglected.
  • Break down the problem: understand and outline each step.
  • Recognize the role of each force and applicable formulas.
  • Anticipate additional factors like air resistance for comprehensive understanding.

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Most popular questions from this chapter

A boxed 10.0 -kg computer monitor is dragged by friction 5.50 \(\mathrm{m}\) up along the moving surface of a conveyor belt inclined at an angle of \(36.9^{\circ}\) above the horizontal. If the monitor's speed is a constant 2.10 \(\mathrm{cm} / \mathrm{s}\) , how much work is done on the monitor by (a) friction, (b) gravity, and (c) the normal force of the conveyor belt?

Working Like a Horse. Your job is to lift 30 -kg crates a vertical distance of 0.90 m from the ground onto the bed of a truck. (a) How many crates would you have to load onto the truck in 1 minute for the average power output you use to lift the crates to equal 0.50 \(\mathrm{hp} ?\) (b) How many crates for an average power output of 100 \(\mathrm{W} ?\)

When its \(75-\mathrm{kW}(100 \mathrm{-hp})\) engine is generating full power, a small single-engine airplane with mass 700 \(\mathrm{kg}\) gains altitude at a rate of 2.5 \(\mathrm{m} / \mathrm{s}(150 \mathrm{m} / \mathrm{min}\) , or 500 \(\mathrm{ft} / \mathrm{min}\) ). What fraction of the engine power is being used to make the airplane climb? (The remainder is used to overcome the effects of air resistance and of inefficiencies in the propeller and engine.)

CALC Consider a spring that does not obey Hooke's law very faithfully. One end of the spring is fixed. To keep the spring stretched or compressed an amount \(x,\) a force along the \(x\) -axis with \(x\) -component \(F_{x}=k x-b x^{2}+c x^{3}\) must be applied to the free end. Here \(k=100 \mathrm{N} / \mathrm{m}, b=700 \mathrm{N} / \mathrm{m}^{2},\) and \(c=12,000 \mathrm{N} / \mathrm{m}^{3} .\) Note that \(x>0\) when the spring is stretched and \(x<0\) when it is compressed. (a) How much work must be done to stretch this spring by 0.050 m from its unstretched length? (b) How much work must be done to compress this spring by 0.050 m from its unstretched length? (c) Is it easier to stretch or compress this spring? Explain why in terms of the dependence of \(F_{x}\) on \(x\) . (Many real springs behave qualitatively in the same way.)

A sled with mass 8.00 \(\mathrm{kg}\) moves in a straight line on a frictionless horizontal surface. At one point in its path, its speed is \(4.00 \mathrm{m} / \mathrm{s} ;\) after it has traveled 2.50 \(\mathrm{m}\) beyond this point, its speed is 6.00 \(\mathrm{m} / \mathrm{s}\) . Use the work-energy theorem to find the force acting on the sled, assuming that this force is constant and that it acts in the direction of the sled's motion.
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