/*! 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 6 A baseball player with mass \(m=... [FREE SOLUTION] | 91Ó°ÊÓ

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A baseball player with mass \(m=79 \mathrm{~kg}\), sliding into second base, is retarded by a frictional force of magnitude \(470 \mathrm{~N}\). What is the coefficient of kinetic friction \(\mu_{k}\) between the player and the ground?

Short Answer

Expert verified
The coefficient of kinetic friction \(\mu_{k}\) is approximately 0.606.

Step by step solution

01

Identify the forces involved

Begin by identifying the forces acting on the player. When sliding, the player experiences a normal force \(N\) from the ground, and a frictional force \(f\) which is acting in the opposite direction of motion. The normal force is equal to the gravitational force acting on the player, which can be calculated as \(mg\), where \(g = 9.81 \, \mathrm{m/s^2}\).
02

Write the formula for frictional force

The frictional force \(f\) is given by the equation \(f = \mu_{k} N\). Since the normal force \(N\) is equal to the gravitational force, \(N = mg\). So, the equation becomes \(f = \mu_{k} mg\).
03

Rearrange the formula to find \(\mu_{k}\)

Rearrange the equation \(f = \mu_{k} mg\) to solve for the coefficient of kinetic friction \(\mu_{k}\): \[ \mu_{k} = \frac{f}{mg} \]
04

Substitute the given values

Substitute the given values into the formula: \(f = 470 \, \mathrm{N}\), \(m = 79 \, \mathrm{kg}\), and \(g = 9.81 \, \mathrm{m/s^2}\). Perform the calculation: \[ \mu_{k} = \frac{470}{79 \times 9.81} \]
05

Calculate \(\mu_{k}\)

Perform the calculation: \[ \mu_{k} = \frac{470}{774.99} \]Simplifying this gives:\[ \mu_{k} \approx 0.606 \]

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Key Concepts

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

Kinetic Friction
Kinetic friction occurs when two surfaces slide against each other. It is an opposing force that resists the motion of the moving body. Imagine running your hand across a table; the force that makes it a little hard to slide your hand is the kinetic friction. This type of friction acts parallel to the surfaces in contact and opposite the direction of movement.
Kinetic friction is characterized by a specific coefficient, known as the coefficient of kinetic friction (\( \mu_k \)), a constant that varies based on the materials involved. This coefficient provides a measure of how "sticky" or "slippery" a surface is when in motion, allowing us to calculate the frictional force, which is crucial in solving many physics problems. The formula for kinetic frictional force is:
  • \( f = \mu_k N \)
where \( N \) is the normal force. This indicates that frictional force depends on both the coefficient of friction and the normal force acting on the sliding object.
Normal Force
The normal force is a supportive force exerted perpendicular to the surfaces in contact. When an object, like our baseball player, is on a flat surface, this force acts upwards, balancing the gravitational force pulling the object downward. It's called "normal" because it is at right angles to the contact surface.In physics problems, especially those involving friction, calculating the normal force is often necessary. For a situation with no vertical acceleration, like our baseball player sliding on level ground, the normal force is equal in magnitude and opposite in direction to the gravitational force, \( mg \). This means the normal force \( N \) is given by:
  • \( N = mg \)
This formula simplifies many friction calculations because it allows for quick determination of how strong the support from the surface is, which is fundamental to finding the frictional force involved.
Coefficient of Friction
The coefficient of friction is a dimensionless scalar value that describes the ratio of the force of friction between two bodies and the force pressing them together. In simpler terms, it measures how much friction force acts relative to the normal force. There are two main types: static and kinetic, but here we're focusing on kinetic friction.The coefficient of kinetic friction (\( \mu_k \)) tells us how rough or smooth the sliding surfaces are. High values mean more friction, requiring more force to slide an object, while low values indicate smoother surfaces that slide more easily. This coefficient is calculated using:
  • \( \mu_k = \frac{f}{N} \)
In our exercise, we rearranged this equation to solve for \( \mu_k \) using known values for frictional force \( f \) and normal force \( N = mg \). This calculation allows us to determine how effectively the player's surfaces interact, giving us an insight into the nature of the materials and contact surfaces involved.

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Most popular questions from this chapter

A circular curve of highway is designed for traffic moving at \(60 \mathrm{~km} / \mathrm{h}\). Assume the traffic consists of cars without negative lift. (a) If the radius of the curve is \(150 \mathrm{~m}\), what is the correct angle of banking of the road? (b) If the curve were not banked, what would be the minimum coefficient of friction between tires and road that would keep traffic from skidding out of the turn when traveling at \(60 \mathrm{~km} / \mathrm{h} ?\)

The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of \(0.25\) with the floor. If the train is initially moving at a speed of \(48 \mathrm{~km} / \mathrm{h}\), in how short a distance can the train be stopped at constant acceleration without causing the crates to slide over the floor?

An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is \(5.0 \mathrm{kN}\), and the circle's radius is \(10 \mathrm{~m}\). At the top of the circle, what are the (a) magnitude \(F_{B}\) and (b) direction (up or down) of the force on the car from the boom if the car's speed is \(v=5.0 \mathrm{~m} / \mathrm{s} ?\) What are (c) \(F_{B}\) and (d) the direction if \(v=12 \mathrm{~m} / \mathrm{s} ?\)

In the early afternoon, a car is parked on a street that runs down a steep hill, at an angle of \(35.0^{\circ}\) relative to the horizontal. Just then the coefficient of static friction between the tires and the street surface is \(0.725 .\) Later, after nightfall, a sleet storm hits the area, and the coefficient decreases due to both the ice and a chemical change in the road surface because of the temperature decrease. By what percentage must the coefficient decrease if the car is to be in danger of sliding down the street?

A child places a picnic basket on the outer rim of a merry-goround that has a radius of \(4.6 \mathrm{~m}\) and revolves once every \(30 \mathrm{~s}\). (a) What is the speed of a point on that rim? (b) What is the lowest value of the coefficient of static friction between basket and merrygo-round that allows the basket to stay on the ride?

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