Lars, of mass \(82.4 \mathrm{kg},\) has been working out and can do work for about 2.0 min at the rate of \(1.0 \mathrm{hp}(746 \mathrm{W})\) How long will it take him to climb three flights of stairs, a vertical height of $12.0 \mathrm{m} ?$

Short Answer

Expert verified
Answer: It takes the student 13 seconds to climb the three flights of stairs, which is within his 2.0-minute ability to work at a rate of 1.0 hp.

Step by step solution

01

Calculate gravitational potential energy gained by the student while climbing the stairs

We can calculate gravitational potential energy using the formula: \(E_p = mgh\), where \(E_p\) is the gravitational potential energy, \(m\) is the mass (82.4 kg), \(g\) is the acceleration due to gravity (9.81 \(m/s^{2}\)), and \(h\) is the height (12.0 m). \(E_p = (82.4 \mathrm{kg})(9.81 \mathrm{m/s^2})(12.0 \mathrm{m}) = 9704.448 \mathrm{J}\) (Joules)
02

Calculate total work the student can do in 2.0 minutes

Since the student can work at a rate of 1.0 hp (746 W) for 2.0 minutes, we first need to convert 2.0 minutes into seconds. There are 60 seconds in a minute, so: Time = \(2.0 \times 60 = 120 \ \mathrm{s}\) (seconds) Now calculate the total work the student can do in 2.0 minutes: Work = Power x Time Work = \(746 \ \mathrm{W} \times 120\ \mathrm{s} = 89520 \mathrm{J}\)
03

Calculate the time it takes the student to climb the stairs

We will use the power formula to determine the time it takes for the student to climb the stairs: Power = Work / Time Rearranging the formula to solve for time: Time = Work / Power Since we found in step 1 that the work needed to climb the stairs (gravitational potential energy gained) is 9704.448 J, and we know the power at which the student works (1.0 hp or 746 W): Time = \(\dfrac{9704.448 \ \mathrm{J}}{746 \ \mathrm{W}} = 13 \ \mathrm{s}\) It will take the student 13 seconds to climb the three flights of stairs, which is within his 2.0-minute ability to work at a rate of 1.0 hp.

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 Vaia!

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

When the spring on a toy gun is compressed by a distance \(x\), it will shoot a rubber ball straight up to a height of \(h .\) Ignoring air resistance, how high will the gun shoot the same rubber ball if the spring is compressed by an amount \(2 x ?\) Assume \(x<<h\)
A \(62-\mathrm{kg}\) woman takes \(6.0 \mathrm{s}\) to run up a flight of stairs. The landing at the top of the stairs is \(5.0 \mathrm{m}\) above her starting place. (a) What is the woman's average power output while she is running? (b) Would that be equal to her average power input-the rate at which chemical energy in food or stored fat is used? Why or why not?
The escape speed from Earth is \(11.2 \mathrm{km} / \mathrm{s},\) but that is only the minimum speed needed to escape Earth's gravitational pull; it does not give the object enough energy to leave the solar system. What is the minimum speed for an object near the Earth's surface so that the object escapes both the Earth's and the Sun's gravitational pulls? Ignore drag due to the atmosphere and the gravitational forces due to the Moon and the other planets. Also ignore the rotation and the orbital motion of the Earth.
The orbit of Halley's comet around the Sun is a long thin ellipse. At its aphelion (point farthest from the Sun), the comet is $5.3 \times 10^{12} \mathrm{m}\( from the Sun and moves with a speed of \)10.0 \mathrm{km} / \mathrm{s} .$ What is the comet's speed at its perihelion (closest approach to the Sun) where its distance from the Sun is \(8.9 \times 10^{10} \mathrm{m} ?\)
Bruce stands on a bank beside a pond, grasps the end of a 20.0 -m-long rope attached to a nearby tree and swings out to drop into the water. If the rope starts at an angle of \(35.0^{\circ}\) with the vertical, what is Bruce's speed at the bottom of the swing?
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.

Sign-up for free