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

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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.

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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

the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

-reverse card

Answer from: Quest

Ithink it’s 98 because jajajjajansmsmjs

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