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Arlecino [84]
2 years ago
12

A block of mass m = 1.00 kg is attached to a spring of force constant k = 500 N/m. The block is pulled to a position xi= 5.00 cm

to the right of equilibrium and released from rest. Find the speed the block has as it passes through equilibrium if the coefficient of friction between block and surface is 0.350.
Physics
1 answer:
8_murik_8 [283]2 years ago
6 0

Answer:

The speed of the block is 4.96 m/s.

Explanation:

Given that.

Mass of block = 1.00 kg

Spring constant = 500 N/m

Position x_{i}=5.00\ cm

Coefficient of friction = 0.350

(A). We need to calculate the speed the block has as it passes through equilibrium if the horizontal surface is friction less

Using formula of kinetic energy and potential energy

\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2-\mu mgx

Put the value into the formula

\dfrac{1}{2}\times1.00\times v^2=\dfrac{1}{2}\times500\times(5.00\times10^{-2})-0.350\times1.00\times9.8\times5.00\times10^{-2}

v^2=\dfrac{2\times12.3285}{1.00}

v^2=24.657

v=4.96\ m/s

Hence, The speed of the block is 4.96 m/s.

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1) draw a simple circuit with a voltage source and four resistors wired in series
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Answer:

1)

In this circuit (see attachment #1), we have:

- A voltage source: in this case, we choose a battery. A voltage source is a device producing an electromotive force (in a battery, this is done by means of a chemical reaction), which is responsible for "pushing" the electrons along the circuit and creating a current. The electromotive force (emf) of the battery is also called voltage, and it is indicated with the letter V.

- Four resistors: a resistor is a device which opposes to the flow of current. The property that describes by "how much" the resistor "opposes" to the flow of current is called "resistance", and it is indicated with the letter R.

- In this circuit, the 4 resistors are in series. Resistors are said to be in series when they are connected along the same branch of the circuit, so that the same current flow across each of them.

- For resistors in series, the equivalent resistance of the circuit is given by the sum of the individual resistances:

R=R_1+R_2+...+R_n

2)

In this circuit (see attachment #2), we have:

- A voltage source: as before, we have chosen a battery, providing an electromotive  force to the circuit

- Three resistors wired in parallel. Resistors are said to be connected in parallel when they are connected along different branches, but with their terminals connected to the same point, so that each of them has the same potential difference across it.

- For resistors in parallel, the equivalent resistance of the circuit is calculated using the formula:

\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+...+\frac{1}{R_n}

3)

In this circuit (see attachment #3), we have:

- A voltage source (again, we have choosen a battery)

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-- 2 of them are connected in parallel with each other

-- the 3rd one it is in series with the first two

If we call R_1,R_2 the resistances of the first 2 resistors in parallel, their equivalent resistance is:

\frac{1}{R_{12}}=\frac{1}{R_1}+\frac{1}{R_2}\\\rightarrow R_{12}=\frac{R_1 R_2}{R_1+R_2}

Then, these two resistors are connected in series with resistor R_3; and so, the total resistance of this circuit will be:

R=R_{12}+R_3=\frac{R_1R_2}{R_1+R_2}+R_3=\frac{R_1R_2+R_3(R_1+R_2)}{R_1+R_2}

4)

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R_{123}=R_1+R_2+R_3

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\frac{1}{R}=\frac{1}{R_{123}}+\frac{1}{R_{456}}\\\rightarrow R=\frac{R_{123}R_{456}}{R_{123}+R_{456}}=\frac{(R_1+R_2+R_3)(R_4+R_5+R_6)}{R_1+R_2+R_3+R_4+R_5+R_6}

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3 years ago
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