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

Short Answer

Expert verified
The 100-watt bulb uses 360,000 joules per hour. A 70-kg person must run at about 101.42 m/s to have the same kinetic energy.

Step by step solution

01

Understanding Energy Use

A watt measures power, which is energy per unit of time. Specifically, a watt equals one joule per second. Thus, a 100-watt bulb uses 100 joules of energy every second.
02

Calculate Energy Used per Hour

To find out how many joules the bulb uses in an hour, we need to calculate the energy use over that period. Since there are 3600 seconds in an hour, multiply the power by the number of seconds: \(100 \text{ watts} \times 3600 \text{ seconds} = 360000 \text{ joules}\).
03

Understanding Kinetic Energy Formula

The formula for kinetic energy (KE) is \( KE = \frac{1}{2} m v^2 \), where \(m\) is mass in kilograms, and \(v\) is velocity in meters per second. This represents the energy an object has due to its motion.
04

Set Up the Equation

We know the kinetic energy must equal the energy the bulb uses in one hour, so set \( KE = 360000 \text{ joules} \). Plug in the person's mass \( m = 70 \text{ kg} \): \( 360000 = \frac{1}{2} \times 70 \times v^2 \).
05

Solve for Velocity

Rearrange the kinetic energy formula to solve for velocity \(v\):\[360000 = 35v^2 \v^2 = \frac{360000}{35} \v^2 \approx 10285.71 \v \approx \sqrt{10285.71} \v \approx 101.42 \] Thus, the velocity is approximately 101.42 meters per second.

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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 an object moves, it has the capacity to do work, which is directly related to its mass and velocity. The formula for kinetic energy is given by: \[ KE = \frac{1}{2} mv^2 \] where:
  • \( KE \) is the kinetic energy
  • \( m \) is the mass of the object in kilograms
  • \( v \) is the velocity of the object in meters per second
This equation shows that kinetic energy increases with the square of the velocity; thus, small increases in velocity can result in large increases in kinetic energy. This principle is crucial in understanding the physical impact of moving objects. For a 70 kg person, to match the energy used by a 100-watt bulb in an hour, they would need to run at a specific velocity. By rearranging the formula, we can solve for the required speed, which provides practical insight into the kinetic energy concept.
power and energy
Power and energy are closely related, yet distinct concepts. Power refers to the rate at which energy is used or transferred, measured in watts (W).
  • One watt equals one joule per second. Hence, power tells us how much energy a device uses per second.
In contrast, energy is the total amount used over time and is measured in joules (J). These concepts are fundamental when analyzing devices like a 100-watt light bulb, which consumes 100 joules every second. Over an hour, this amounts to a significant total energy use, calculated by multiplying the power by time (in seconds). Understanding how power relates to energy helps us manage electricity usage and improve efficiency. Daily, it gives rise to practical applications such as measuring the total electricity consumed in households. By evaluating wattage and usage time, one can estimate their energy consumption and potentially reduce costs through more efficient energy practices.
joules to watt-hour conversion
The conversion between joules and watt-hours is an essential skill for understanding energy usage in practical terms. Since many devices state energy consumption in kilowatt-hours (kWh), comprehending how to convert from joules helps you grasp this measurement.To convert joules to watt-hours, understand that:
  • 1 watt-hour equals 3600 joules, as one watt is one joule/second and there are 3600 seconds in an hour.
Therefore, to determine how many watt-hours corresponds to a certain number of joules, divide the total joules by 3600.For instance, a 100-watt bulb that uses 360,000 joules in an hour is equivalent to \(\frac{360000}{3600} = 100\) watt-hours. This method helps you compare energy consumption across various appliances easily, especially when evaluating cost or efficiency based on household power bills. Making this conversion part of everyday considerations can aid in actionable improvements toward more sustainable energy consumption habits.

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Most popular questions from this chapter

Leg Presses. As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arranged side by side so that they are parallel to each other. When you push the platform, you compress the springs. You do 80.0 J of work when you compress the springs 0.200 \(\mathrm{m}\) from their uncompressed length.(a) What magnitude of force must you apply to hold the platform in this position? (b) How much additional work must you do to move the platform 0.200 \(\mathrm{m}\) farther, and what maximum force must you apply?

A 5.00 -kg block is moving at \(v_{0}=6.00 \mathrm{m} / \mathrm{s}\) along a frictionless, horizontal surface toward a spring with force constant \(k=500 \mathrm{N} / \mathrm{m}\) that is attached to a wall (Fig. P6.85). The spring has negligible mass. \begin{equation} \begin{array}{l}{\text { (a) Find the maximum distance the spring will be compressed. }} \\ {\text { (b) If the spring is to compress by no more than } 0.150 \mathrm{m}, \text { what }} \\\ {\text { should be the maximum value of } v_{0} ?}\end{array} \end{equation}

A factory worker pushes a 30.0 -kg crate a distance of 4.5 \(\mathrm{m}\) along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is 0.25 . (a) What magnitude of force must the worker apply? (b) How much work is done on the crate by this force? (c) How much work is done on the crate by friction? (d) How much work is done on the crate by the normal force? By gravity? (e) What is the total work done on the crate?

Two blocks are connected by a very light string passing over a massless and frictionless pulley (Fig. E6.7). Traveling at constant speed, the 20.0 -N block moves 75.0 \(\mathrm{cm}\) to the right and the 12.0 -N block moves 75.0 \(\mathrm{cm}\) downward. During this process, how much work is done (a) on the \(12.0-\mathrm{N}\) block by (i) gravity and (ii) the tension in the string? (b) On the 20.0 -N block by (i) gravity, (ii) the tension in the string, (iii) friction, and (iv) the normal force? (c) Find the total work done on each block.

Meteor Crater. About \(50,000\) years ago, a meteor crashed into the earth near present-day Flagstaff, Arizona. Measurements from 2005 estimate that this meteor had a mass of about \(1.4 \times 10^{8}\) kg (around \(150,000\) tons) and hit the ground at a speed of 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 amount of energy as a million tons of TNT, and 1.0 ton of TNT releases \(4.184 \times 10^{9}\) J of energy.)

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