/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 51 How many joules of energy does a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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
A 100-watt bulb uses 360,000 joules per hour. A 70 kg person must run at 101.42 m/s to match that energy in kinetic form.

Step by step solution

01

Convert Watts to Joules

To find how many joules the bulb uses per hour, we start with its power rating: 100 watts. Since 1 watt = 1 joule/second, a 100-watt bulb uses \(100 \text{ joules/second}\). To find the energy consumption in one hour, multiply by the number of seconds in an hour (3600):\[100 \text{ watts} \times 3600 \text{ seconds} = 360,000 \text{ joules}.\]
02

Kinetic Energy Formula

The kinetic energy \(E_k\) of an object is given by the formula \(E_k = \frac{1}{2}mv^2\), where \(m\) is mass and \(v\) is velocity. We need to find the velocity that gives a 70 kg person 360,000 joules of energy.
03

Solve for Velocity

Rearrange the kinetic energy formula to solve for velocity \(v\): \[v = \sqrt{\frac{2E_k}{m}}.\] Substitute \(E_k = 360,000\) joules and \(m = 70\) kg: \[v = \sqrt{\frac{2 \times 360,000}{70}}.\] Calculate: \[v \approx \sqrt{10,285.71} \approx 101.42 \text{ m/s}.\]
04

Conclusion

A 70 kg person needs to run at approximately 101.42 meters per second to have 360,000 joules of kinetic energy, the same as the energy used by a 100-watt light bulb in one hour.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Energy Conversion
Energy conversion is the process of transforming energy from one form to another. In our exercise, we focus on converting electrical power, measured in watts, into energy, measured in joules. This is a common conversion since electrical appliances, like our 100-watt light bulb, have power ratings that help us understand their energy usage. Understanding energy conversion allows us to compare different forms of energy and see how they relate in practical applications.
Watts to Joules
Converting watts to joules can be straightforward once you know the basics. Watts measure power, which is the rate of energy use or generation. Joules, on the other hand, measure energy itself. One watt is equivalent to one joule per second.

To convert watts to joules for any duration, you multiply the watts by the number of seconds during which the energy flow is happening. So, in our example:
  • The light bulb is 100 watts, meaning it uses 100 joules every second.
  • There are 3600 seconds in an hour.
  • Therefore, in one hour, the bulb uses 100 watts × 3600 seconds, equating to 360,000 joules.
This practical conversion process is key for understanding energy consumption in everyday devices.
Mechanical Energy
Mechanical energy is the sum of kinetic and potential energy in an object related to its motion and position. In this exercise, we are mainly concerned with kinetic energy.

Kinetic energy is the energy possessed due to motion and can be calculated using the formula: \[ E_k = \frac{1}{2}mv^2 \]
  • \( m \) is mass
  • \( v \) is velocity.
In our scenario with the person running, we focus entirely on kinetic energy to achieve the equivalent of the light bulb's energy. The more massive or faster an object, the greater its kinetic energy.

Understanding mechanical energy helps us appreciate how energy is harnessed and exerted, whether in machinery or living organisms.
Velocity Calculation
Calculation of velocity is essential when determining how fast an object must travel to possess a certain amount of kinetic energy. To solve for velocity in the context of kinetic energy, we rearrange the kinetic energy formula:\[ v = \sqrt{\frac{2E_k}{m}} \]In our exercise, \( E_k \) is 360,000 joules, and \( m \) is 70 kg.Substitute these values into the equation to find:\[ v = \sqrt{\frac{2 \times 360,000}{70}} \]This calculation gives us a velocity of approximately 101.42 meters per second.

This process shows how changes in kinetic energy reflect different speeds needed based on mass. It highlights the relationship between velocity and energy, illustrating that even small increases in speed can significantly impact energy in motion.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two blocks are connected by a very light string passing over a massless and frictionless pulley (\(\textbf{Fig. E6.7}\)). Traveling at constant speed, the 20.0-N block moves 75.0 cm to the right and the 12.0-N block moves 75.0 cm downward. How much work is done (a) on the 12.0-N block by (i) gravity and (ii) the tension in the string? (b) How much work is done 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.

A physics professor is pushed up a ramp inclined upward at 30.0\(^\circ\) above the horizontal as she sits in her desk chair, which slides on frictionless rollers. The combined mass of the professor and chair is 85.0 kg. She is pushed 2.50 m along the incline by a group of students who together exert a constant horizontal force of 600 N. The professor's speed at the bottom of the ramp is 2.00 m/s. Use the work\(-\)energy theorem to find her speed at the top of the ramp.

You apply a constant force \(\overrightarrow{F} =(-68.0 \, \mathrm{N})\hat{\imath} +(36.0 \, \mathrm{N})\hat{\jmath}\) to a 380-kg car as the car travels 48.0 m in a direction that is 240.0\(^\circ\) counterclockwise from the +\(x\)-axis. How much work does the force you apply do on the car?

While hovering, a typical flying insect applies an average force equal to twice its weight during each downward stroke. Take the mass of the insect to be 10 g, and assume the wings move an average downward distance of 1.0 cm during each stroke. Assuming 100 downward strokes per second, estimate the average power output of the insect.

A 128.0-N carton is pulled up a frictionless baggage ramp inclined at 30.0\(^\circ\) above the horizontal by a rope exerting a 72.0-N pull parallel to the ramp's surface. If the carton travels 5.20 m along the surface of the ramp, calculate the work done on it by (a) the rope, (b) gravity, and (c) the normal force of the ramp. (d) What is the net work done on the carton? (e) Suppose that the rope is angled at 50.0\(^\circ\) above the horizontal, instead of being parallel to the ramp's surface. How much work does the rope do on the carton in this case?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.