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19 days ago

6

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 fi

nd 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

1 answer:

Sav [2.2K]19 days ago

8 0

Answer:

b = 0.6487 kg / s

Explanation:

In the context of oscillatory motion, friction is related to velocity,

fr = - b v

where b represents the friction coefficient.

Upon solving the equation, the angular velocity is represented as

w² = k / m - (b / 2m)²

In this case, we're given an angular frequency w = 1Hz, the mass m = 0.1 kg, and the spring constant k = 5 N / m. This allows us to derive the friction coefficient.

Let’s denote

w₀² = k / m

w² = w₀² - b² / 4m²

b² = (w₀² -w²) 4 m²

Now, let's calculate the angular frequencies.

w₀² = 5 / 0.1

w₀² = 50

w = 2π f

w = 2π 1

w = 6.2832 rad / s

Substituting values yields

b² = (50 - 6.2832²) 4 0.1²

b = √ 0.42086

b = 0.6487 kg / s

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This assertion is inaccurate.

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15 days ago

A 7.5 kg cannon ball leaves a canon with a speed of 185 m/s. Find the average net force applied to the ball if the cannon muzzle

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x = 0.5*a*t^2

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22 days ago

It's a snowy day and you're pulling a friend along a level road on a sled. You've both been taking physics, so she asks what you

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Answer:

0.0984

Explanation:

The first diagram below illustrates a free body diagram that will aid in resolving this problem.

According to the diagram, the force's horizontal component can be expressed as:

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Substituting 42° for θ and 87.0° for $F$

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Meanwhile, the vertical component is:

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In resolving the vector, let A denote the components in mutually perpendicular directions.

The magnitudes of both components are illustrated in the second diagram provided and can be represented as A cos θ and A sin θ

The frictional force can be expressed as:

$f = \mu \ N$

Where;

$\mu$ is the coefficient of friction

N = the normal force

Also, the normal reaction (N) is calculated as mg - F sin θ

Substituting $F_Y \ for \ F_{sin \ \theta}$ . Normal reaction becomes:

$N = mg \ - \ F_Y$

By balancing the forces, the horizontal component of the force equals the frictional force.

The horizontal component is described as follows:

$F_X = \mu \ ( mg - \ F_Y)$

Rearranging the equation above to isolate $\mu$ leads to:

$\mu \ = \ \frac{F_X}{mg - F_Y}$

Substituting in the following values:

$F_X \ = \ 64.65 \ N$

m = 73 kg

g = 9.8 m/s²

$F_Y = \ 58.21 N$

Thus:

$\mu \ = \ \frac{64.65 N}{(73.0 kg)(9.8m/s^2) - (58.21 \ N)}$

$\mu = 0.0984$

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27 days ago

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the temperature on the left side is 1.48 times greater than that on the right

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similarly

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11 days ago

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Answer:

$\Delta T = 0.81 ^oC$

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