Physics, 20.08.2020 23:01 melkumathurin

Exercise 2.4.6: Suppose you wish to measure the friction a mass of 0.1 kg experiences as it slides along a floor (you wish to find c). You have a spring with spring constant k 5 N/m. You take the spring, you attach it to the mass and fix it to a wall. Then you pull on the spring and let the mass go. You find that the mass oscillates with frequency 1 Hz. What is the friction

Answers: 3

Show answers

Answers

Answer from: yuluvk

The coefficient friction of the mass during the measurement is 0.648 kg/s.

The given parameters;

mass, m = 0.1 kgspring constant, k = 5 N/mfrequency of the mass, F = 1 Hz

During oscillatory motion, friction is directly proportional to speed.

F_k = -vb

where;

b is the coefficient of friction

The angular velocity is given as;

$\omega ^2 = \frac{k}{m} - \frac{b^2}{4m^2} \\\\\omega ^2 = \omega _0^2 - \frac{b^2}{4m^2}\ \ ---\ (1)$

From the equation above, we will have the following;

$\omega_0^2 = \frac{k}{m} \\\\\omega_0^2 = \frac{5}{0.1} \\\\\omega_0^2 = 50$

Also, the instantaneous angular speed is calculated as;

$\omega = 2\pi f\\\\\omega = 2\pi \times 1\\\\\omega = 2\pi\\\\\omega = 6.284 \ rad/s$

From equation (1), the coefficient of friction is calculated as follows;

$\omega ^2 = \omega ^2_0 - \frac{b^2}{4m^2} \\\\ \frac{b^2}{4m^2} = \omega ^2_0 - \omega ^2 \\\\b^2 = 4m^2( \omega ^2_0 - \omega ^2)\\\\b= \sqrt{ 4m^2( \omega ^2_0 - \omega ^2)}\\\\b = \sqrt{ 4\times 0.1^2\times ( 50 - 6.284^2)}\\\\b = 0.648 \ \ kg/s$

Thus, the coefficient friction of the mass during the measurement is 0.648 kg/s.

Learn more here:link

Answer from: mullanebrianot3dpw

b = 0.6487 kg / s

Explanation:

In an oscillatory motion, friction is proportional to speed,

fr = - b v

where b is the coefficient of friction

when solving the equation the angular velocity has the form

w² = k / m - (b / 2m)²

In this exercise we are given the angular velocity w = 1Hz, the mass of the body m = 0.1 kg, and the spring constant k = 5 N / m. Therefore we can disperse the coefficient of friction

let's call

w₀² = k / m

w² = w₀² - b² / 4m²

b² = (w₀² -w²) 4 m²

Let's find the angular velocities

w₀² = 5 / 0.1

w₀² = 50

w = 2π f

w = 2π 1

w = 6.2832 rad / s

we subtitute

b² = (50 - 6.2832²) 4 0.1²

b = √ 0.42086

b = 0.6487 kg / s

Answer from: Quest

what type of charge does it have? negative

Answer from: Quest

the body's situation won't change the forces are equal opposit and on the same body

Another question on Physics

Physics, 21.06.2019 20:30

Air enters a compressor operating at steady state at 1.05 bar, 300 k, with a volumetric flow rate of 93 m3/min and exits at 12 bar, 400 k. heat transfer occurs at a rate of 15.5 kw from the compressor to its surroundings. assuming the ideal gas model for air neglecting kinetic potential energy effects, determine the power in put in kw.

Answers: 2

Answer

Physics, 22.06.2019 10:00

Your town is considering building a biodiesel power plant describe at least two advantages and two disadvantages

Answers: 1

Answer

Physics, 22.06.2019 11:00

The charge on an electron is 1.60218 × 10−19 c and its mass is 9.10939 × 10−31 kg. how strong a magnetic field must be experienced by the electron if its path is a circle of radius 5.5 cm? answer in units of t.

Answers: 3

Answer

Physics, 22.06.2019 18:30

Both speed and velocity measure how fast something is moving. however, since speed is not a it does not require a(n question 6 options: a) quantity , unit b) magnitude, unit c) scalar, direction d) vector, direction

Answers: 2

Answer

You know the right answer?

Exercise 2.4.6: Suppose you wish to measure the friction a mass of 0.1 kg experiences as it slides a...

Questions