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Nata [24]
3 years ago
7

A block with mass 0.5 kg is forced against a horizontal spring of negligible mass, compressing the spring a distance of 0.2 m. W

hen released, the block moves on a horizontal tabletop for 1.0m before coming to rest. The force constant k is 100 N/m. What is the coefficient of kinetic friction between the block and tabletop
Physics
1 answer:
lianna [129]3 years ago
5 0

Answer:

So coefficient of kinetic friction will be equal to 0.4081

Explanation:

We have given mass of the block m = 0.5 kg

The spring is compressed by length x = 0.2 m

Spring constant of the sprig k = 100 N/m

Blocks moves a horizontal distance of s = 1 m

Work done in stretching the spring is equal to W=\frac{1}{2}kx^2=\frac{1}{2}\times 100\times 0.2^2=2J

This energy will be equal to kinetic energy of the block

And this kinetic energy must be equal to work done by the frictional force

So \mu mg\times s=2

\mu\times  0.5\times 9.8\times 1=2

\mu =0.4081

So coefficient of kinetic friction will be equal to 0.4081

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An aluminum wire having a cross-sectional area equal to 2.20 10-6 m2 carries a current of 4.50 A. The density of aluminum is 2.7
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Answer:

The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

Explanation:

We can find the drift speed by using the following equation:

v = \frac{I}{nqA}

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n: is the number of electrons

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We need to find the number of electrons:

n = \frac{6.022\cdot 10^{23} atoms}{1 mol}*\frac{1 mol}{26.982 g}*\frac{2.70 g}{1 cm^{3}}*\frac{(100 cm)^{3}}{1 m^{3}} = 6.03 \cdot 10^{28} atom/m^{3}                  

Now, we can find the drift speed:

v = \frac{I}{nqA} = \frac{4.50 A}{6.03 \cdot 10^{28} atom/m^{3}*1.6 \cdot 10^{-19} C*2.20 \cdot 10^{-6} m^{2}} = 2.12 \cdot 10^{-4} m/s              

Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

I hope it helps you!      

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2 years ago
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