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2 years ago

PLEASE HURRY

1 answer:

5 0

Decompose the forces acting on the block into components that are parallel and perpendicular to the ramp. (See attached free body diagram. Forces are not drawn to scale)

• The net force in the parallel direction is

∑ F (para) = -mg sin(21°) - f = ma

• The net force in the perpendicular direction is

∑ F (perp) = n - mg cos(21°) = 0

Solving the second equation for n gives

n = mg cos(21°)

n = (0.200 kg) (9.80 m/s²) cos(21°)

n ≈ 1.83 N

Then the magnitude of friction is

f = µn

f = 0.25 (1.83 N)

f ≈ 0.457 N

Solve for the acceleration a :

-mg sin(21°) - f = ma

a = (-0.457N - (0.200 kg) (9.80 m/s²) sin(21°))/(0.200 kg)

a ≈ -5.80 m/s²

so the block is decelerating with magnitude

a = 5.80 m/s²

down the ramp.

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3 years ago

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dmitriy555 [2]

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