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Chapter 6: Problem 43

How many joules of energy does a 100 -watt light bulb use per hour? How fast would a \(70-\mathrm{kg}\) person have to run to have that amount of kinetic energy?

Short Answer

Expert verified
The light bulb uses 360,000 joules per hour; a 70 kg person needs to run approximately 101.42 m/s to have the same kinetic energy.

Step by step solution

01

Understanding the Problem

We need to find the energy consumption of a 100-watt light bulb in joules per hour and then determine the speed a 70-kg person would need to have to possess the same amount of kinetic energy.
02

Calculating Energy Usage

Energy in joules can be calculated from power (watts) and time (seconds) using the formula: \( ext{Energy} = ext{Power} imes ext{Time} \). Since 1 watt = 1 joule/second and 1 hour = 3600 seconds, we calculate: \( 100 ext{ watts} imes 3600 ext{ seconds} = 360,000 ext{ joules} \).
03

Applying the Kinetic Energy Formula

The kinetic energy (KE) of a moving object is given by the formula: \( KE = \frac{1}{2}mv^2 \), where \( m \) is mass and \( v \) is velocity. Here, the person has a mass \( m = 70 ext{ kg} \). We need to find the speed \( v \) that results in 360,000 joules of energy.
04

Solving for Velocity

Rearrange the kinetic energy formula to solve for velocity: \( v^2 = \frac{2 imes KE}{m} \). Substitute \( KE = 360,000 \) joules and \( m = 70 ext{ kg} \) into the equation: \( v^2 = \frac{2 imes 360,000}{70} \). This simplifies to \( v^2 = 10,285.71 \). Take the square root to find \( v \): \( v = \sqrt{10,285.71} \approx 101.42 ext{ m/s} \).

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Key Concepts

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

Joules Calculation
The concept of joules is central to understanding energy. A joule is the standard unit of energy in the International System of Units (SI). Calculating energy in joules involves understanding the relationship between power and time.

To find out how many joules of energy a 100-watt light bulb uses per hour, we use the formula:\[\text{Energy} = \text{Power} \times \text{Time}\]
  • Power is measured in watts, where 1 watt equals 1 joule per second.
  • Time is typically in seconds when using this formula.
For the conversion:
  • A 100-watt light bulb represents 100 joules per second.
  • There are 3600 seconds in an hour, so we multiply 100 watts by 3600 seconds to get 360,000 joules.
This conversion shows how energy used at a constant rate of power is calculated.
Power and Energy
Energy and power are related but distinct concepts. Power is the rate at which energy is used or transferred. It's measured in watts (\(W\)), where one watt is equivalent to one joule per second.

When we talk about a 100-watt light bulb, we refer to its power usage. The bulb uses 100 joules of energy every second to produce light.

Here's a breakdown:
  • Power indicates how quickly energy is being consumed.
  • Energy, measured in joules, tells us the amount of work done or heat generated.
Understanding this relationship is crucial, as it allows us to calculate the total energy consumed over a period, such as one hour, by multiplying the power (in watts) by the time (in seconds). This is why a 100-watt bulb consumes 360,000 joules in an hour.
Kinetic Energy Formula
The kinetic energy formula is important when examining how energy manifests in moving objects. It highlights the relationship between mass, velocity, and kinetic energy.

The formula for kinetic energy (\( KE \)) is:\[KE = \frac{1}{2}mv^2\]
  • \( m \) represents mass in kilograms.
  • \( v \) is the velocity in meters per second.
For example, to determine the speed a 70-kg person needs to run to match the 360,000 joules used by a light bulb:
  • We rearrange the formula to solve for velocity: \( v^2 = \frac{2 \times KE}{m} \)
  • Substitute \( KE = 360,000 \) joules and \( m = 70 \) kg into the equation to find \( v^2 = 10,285.71 \).
  • The velocity \( v \) is about 101.42 meters per second after taking the square root.
This comparison helps illustrate how the energy used by everyday objects compares to energy in motion.

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Most popular questions from this chapter

A pump is required to lift 800 \(\mathrm{kg}\) of water (about 210 gallons) per minute from a well 14.0 \(\mathrm{m}\) deep and eject it with a speed of 18.0 \(\mathrm{m} / \mathrm{s}\) , (a) How much work is done per minute in lifting the water? (b) How much work is done in giving the water the kinetic energy it has when ejected? (c) What must be the power output of the pump?

A Spring with Mass. We usually ignore the kinetic energy of the moving coils of a spring, but let's try to get a reasonable approximation to this. Consider a spring of mass \(M\) , equilibrium length \(L_{0}\) , and spring constant \(k\) . The work done to stretch or compress the spring by a distance \(L\) is \(\frac{1}{2} k X^{2}\) , where \(X=L-L_{0}\) (a) Consider a spring, as described above, that has one end fixed and the other end moving with speed \(v\) . Assume that the speed of points along the length of the spring varies linearly with distance \(l\) from the fixed end. Assume also that the mass \(M\) of the spring is distributed uniformly along the length of the spring. Calculate the kinetic energy of the spring in terms of \(M\) and \(v\) . (Hint: Divide the spring into pieces of length \(d l ;\) find the speed of each piece in terms of \(l, v,\) and \(L ;\) find the mass of each piece in terms of \(d l, M,\) and \(L ;\) and integrate from 0 to \(L\) . The result is not \(\frac{1}{2} M v^{2},\) since not all of the spring moves with the same speed.) In a spring gun, a spring of mass 0.243 \(\mathrm{kg}\) and force constant 3200 \(\mathrm{N} / \mathrm{m}\) is compressed 2.50 \(\mathrm{cm}\) from its unstretched length. When the trigger is pulled, the spring pushes horizontally on a 0.053 -kg ball. The work done by friction is negligible. Calculate the ball's speed when the spring reaches its uncompressed length (b) ignoring the mass of the spring and (c) including, using the results of part (a), the mass of the spring. (d) In part (c), what is the final kinetic energy of the ball and of the spring?

Meteor Crater. About \(50,000\) years ago, a meteor crashed into the earth near present-day Flagstaff, Arizona. Recent \((2005)\) measurements estimate that this meteor had a mass of about \(1.4 \times 10^{3} \mathrm{kg}\) (around \(150,000\) tons) and hit the ground at 12 \(\mathrm{km} / \mathrm{s}\). (a) How much kinetic energy did this meteor deliver to the ground? (b) How does this energy compare to the energy released by a 1.0 -megaton nuclear bomb? (A megaton bomb releases the same energy as a million tons of TNT, and 1.0 ton of TNT releases \(4.184 \times 10^{9} \mathrm{J}\) of energy.)

When its \(75-k W(100-h p)\) engine is generating full power, a small single- engine airplane with mass 700 kg gains altitude at a rate of 2.5 \(\mathrm{m} / \mathrm{s}(150 \mathrm{m} / \mathrm{min} \text { , or } 500 \mathrm{ft} / \mathrm{min})\) . 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.)

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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