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defon
3 years ago
7

A 2.2-kg block slides on a horizontal surface with a speed of v=0.80m/s and an

Physics
1 answer:
mina [271]3 years ago
5 0

Answer:

μ = 0.33

Equal to 3.2 m/s²

Explanation:

Draw a free body diagram of the block.  There are three forces:

Normal force N pushing up.

Weight force mg pulling down.

Friction force Nμ pushing opposite the direction of motion.

Sum of forces in the y direction.

∑F = ma

N − mg = 0

N = mg

Sum of forces in the x direction.

∑F = ma

Nμ = ma

Substitute.

mgμ = ma

μ = a/g

μ = (3.2 m/s²) / (9.8 m/s²)

μ = 0.33

As found earlier, the acceleration is a = gμ.  Since g and μ are constant, a is also constant, so it does not change with velocity.

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a) We have equation of motion

             S = ut + 0.5at²

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          S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m

   Distance traveled during the last second of motion before hitting the ground = 33.45 m

       

3 0
3 years ago
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