Calculate the kinetic energies of the following objects moving at the given speeds: ( \(a\) ) a \(110-\mathrm{kg}\) football linebacker running at \(8.1 \mathrm{~m} / \mathrm{s} ;\) (b) a \(4.2-\mathrm{g}\) bullet at \(950 \mathrm{~m} / \mathrm{s} ;\) ( \(c\) ) the aircraft carrier Nimitz, 91,400 tons at \(32.0\) knots.

Short Answer

Expert verified
The kinetic energy of the linebacker is approximately \(3614.5 \, J\), the bullet approximately \(1898.25 \, J\), and the aircraft carrier approximately \(2.287 \times 10^{11} \, J\).

Step by step solution

01

Calculate the Kinetic Energy of the Linebacker

First, convert the mass of the linebacker to kg, if not already in this unit. In this case, it is. Then, plug the values into the kinetic energy formula to solve for the linebacker's kinetic energy: \(KE = \frac{1}{2}mu^2 = \frac{1}{2}(110 \, kg)(8.1 \, m/s)^2\)
02

Calculate the Kinetic Energy of the Bullet

The mass of the bullet is given in grams, so it needs to be converted to kilograms by dividing by 1,000. After that, calculate the kinetic energy of the bullet by plugging the values into the kinetic energy formula: \(KE = \frac{1}{2}mu^2 = \frac{1}{2}(0.0042 \, kg)(950 \, m/s)^2.\)
03

Calculate the Kinetic Energy of the Aircraft Carrier

First, convert the mass of the aircraft carrier from tons to kilograms by multiplying by 1,000. The speed is given in knots, it needs to be converted to meters per second by multiplying by 0.51444 (since 1 knot = 0.51444 m/s). Now with values in the appropriate units, you can solve for the kinetic energy with the formula: \(KE = \frac{1}{2}mu^2 = \frac{1}{2}(91400000 \, kg)(32 \times 0.51444 \, m/s)^2\)

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

A worker pushed a \(58.7\) -lb block \((m=26.6 \mathrm{~kg})\) a distance of \(31.3 \mathrm{ft}(=9.54 \mathrm{~m})\) along a level floor at constant speed with a force directed \(32.0^{\circ}\) below the horizontal. The coefficient of kinetic friction is \(0.21\). How much work did the worker do on the block?

How much power, in horsepower, must be developed by the engine of a \(1600-\mathrm{kg}\) car moving at \(26 \mathrm{~m} / \mathrm{s}(=94 \mathrm{~km} / \mathrm{h})\) on a level road if the forces of resistance total \(720 \mathrm{~N}\) ?

Delivery trucks that operate by making use of energy stored in a rotating flywheel have been used in Europe. The trucks are charged by using an electric motor to get the flywheel up to its top speed of \(624 \mathrm{rad} / \mathrm{s}\). One such flywheel is a solid, homogeneous cylinder with a mass of \(512 \mathrm{~kg}\) and a radius of \(97.6 \mathrm{~cm} .\) (a) What is the kinetic energy of the flywheel after charging? ( \(b\) ) If the truck operates with an average power requirement of \(8.13 \mathrm{~kW}\), for how many minutes can it operate between chargings?

Two vectors, \(\overrightarrow{\mathbf{r}}\) and \(\overrightarrow{\mathbf{s}}\), lie in the \(x y\) plane. Their magnitudes are \(4.5\) and \(7.3\) units, respectively, whereas their directions are \(320^{\circ}\) and \(85^{\circ}\) measured counterclockwise from the positive \(x\) axis. What is the value of \(\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{s}}\) ?

A swimmer moves through the water at a speed of \(0.22 \mathrm{~m} / \mathrm{s}\). The drag force opposing this motion is \(110 \mathrm{~N}\). How much power is developed by the swimmer?

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