Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 12

A baseball player (Fig. 5-31) with mass \(79 \mathrm{~kg}\), sliding into a base, is slowed by a force of friction of \(470 \mathrm{~N}\). What is the coefficient of kinetic friction between the player and the ground?

Short Answer

Expert verified
The coefficient of kinetic friction between the player and the ground is approximately 0.61.

Step by step solution

01

Calculate the normal force (Fn)

The normal force (Fn) is the weight of the baseball player which can be calculated by multiplying mass with gravity. Here, the mass is \(79 \mathrm{~kg}\) and gravity is \(9.8 \mathrm{~m/s^2}\). So, \(Fn = mass \times gravity = 79 \mathrm{~kg} \times 9.8 \mathrm{~m/s^2} = 774.2 \mathrm{~N}\)
02

Calculate the coefficient of kinetic friction

Now the coefficient of kinetic friction can be calculated with the formula \(\mu = F_f / F_n \), where the frictional force \(F_f\) is given as \(470 \mathrm{~N}\). So, \(\mu = 470 \mathrm{~N} / 774.2 \mathrm{~N} = 0.6072\)

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.

Start your free trial

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

Assume that the standard kilogram would weigh exactly \(9.80 \mathrm{~N}\) at sea level on the equator if the Earth did not rotate. Then take into account the fact that the Earth does rotate, so that this object moves in a circle of radius \(6370 \mathrm{~km}\) (the Earth's radius) in one day. (a) Determine the centripetal force needed to keep the standard kilogram moving in its circular path. ( \(b\) ) Find the force exerted by the standard kilogram on a spring balance from which it is suspended at the equator (its apparent weight).

A \(2400-\mathrm{lb}(=10.7-\mathrm{kN})\) car traveling at \(30 \mathrm{mi} / \mathrm{h}(=13.4 \mathrm{~m} / \mathrm{s})\) attempts to round an unbanked curve with a radius of \(200 \mathrm{ft}\) \((=61.0 \mathrm{~m}) .(a)\) What force of friction is required to keep the car on its circular path? ( \(b\) ) What minimum coefficient of static friction between the tires and road is required?

The coefficient of static friction between the tires of a car and a dry road is \(0.62\). The mass of the car is \(1500 \mathrm{~kg}\). What maximum braking force is obtainable \((a)\) on a level road and \((b)\) on an \(8.6^{\circ}\) downgrade?

An elevator weighing \(6200 \mathrm{lb}\) is pulled upward by a cable with an acceleration of \(3.8 \mathrm{ft} / \mathrm{s}^{2}\). (a) What is the tension in the cable? (b) What is the tension when the elevator is accelerating downward at \(3.8 \mathrm{ft} / \mathrm{s}^{2}\) but is still moving upward?

A student wants to determine the coefficients of static friction and kinetic friction between a box and a plank. She places the box on the plank and gradually raises one end of the plank. When the angle of inclination with the horizontal reaches \(28.0^{\circ}\), the box starts to slip and slides \(2.53 \mathrm{~m}\) down the plank in \(3.92 \mathrm{~s}\). Find the coefficients of friction.
See all solutions

Recommended explanations on Physics Textbooks

Magnetism and Electromagnetic Induction

Read Explanation

Dynamics

Read Explanation

Physics of Motion

Read Explanation

Electromagnetism

Read Explanation

Astrophysics

Read Explanation

Geometrical and Physical Optics

Read Explanation
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