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A 45.8 kg block is placed on an inclined plane that is 44.2 degrees from the horizontal. What is the acceleration of the block? Ignore friction.I have to do the following:1. Draw a free body diagram2.Identify Givens and Unknowns3.Identify the Equations4.Set up the equation using the givens and unknowns5.Solve

Sagot :

Answer:

The acceleration of the block = 6.83 m/s²

Explanation:

The free body diagram representing the given description is:

The mass of the block, m = 45.8 kg

The angle of inclination, θ = 44.2°

Acceleration due to gravity, g = 9.8 m/s²

Applying Newton's second law of motion

∑F = ma

mgsinθ = ma

Divide both sides by m

[tex]\begin{gathered} \frac{mg\sin \theta}{m}=\text{ }\frac{ma}{m} \\ a\text{ = gsin}\theta \end{gathered}[/tex]

a = 9.8 sin (44.2)

a = 9.8(0.697)

a = 6.83 m/s²

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