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const2013 [10]
4 years ago
13

A 1.4-kg block slides freely across a rough surface such that the block slows down with an acceleration of â1.25 m/s2. what is t

he coefficient of kinetic friction between the block and the surface?
Physics
1 answer:
Marianna [84]4 years ago
6 0

Mass of the block = 1.4 kg

Weight of the block = mg = 1.4 × 9.8 = 13.72 N

Normal force from the surface (N) = 13.72 N

Acceleration = 1.25 m/s^2

Let the coefficient of kinetic friction be μ

Friction force = μN

F(net) = ma

μmg = ma

μg = a

μ = \frac{a}{g}

μ = \frac{1.25}{9.8}

μ = 0.1275

Hence, the coefficient of kinetic friction is: μ = 0.1275

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If 270 watts of power is used in 42 seconds, how much work was done<br>​
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Answer: W = 11340J

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Hey there! I will give the following steps, if you have any questions feel free to ask me in the comments below.

So this is the Formula: Power = Work / Time.

<u>Step 1:</u><em><u> Find the Formula</u></em>

P = W / T

<em><u> </u></em>

<u>Step 2: </u><u><em>Make W the subject of the equation.</em></u>

W = PT

<u>Step 3:</u><u> </u><u><em>Given.</em></u>

P = 270 Watts, T = 42 seconds

<u>Step 4:</u><u><em> Substitute these values into equation 2 .</em></u>

W = 270(42)

<u>Step 5:</u><u> </u><u><em>Simplify.</em></u>

W = 11340J

The amount of work done was 11340.

~I hope I helped you! :)~

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