A 45 kg crate is sliding down an incline with an angle of 55 degrees to the horizontal. The coefficient of friction is 0.24. a. Draw a freebody diagram. b. Determine the weight of the crate. c. Determine the component of weight parallel to the plane d. Determine the Normal force. e. Determine the frictional force f. Determine the acceleration of the crate.

Using physics formulas and a firm understanding of vectors, we calculated the various forces acting upon a crate sliding down a plane. This involved the application of gravity, normal force, friction, and the resultant acceleration of the crate.

Explanation:

a. A freebody diagram would show the block in the middle, with arrows pointing down for the force of gravity, up for the normal force, sideways (along the plane) for the force due to gravity, and in the opposite direction for friction.

b. To determine the weight of the crate, we multiply the mass of the crate by the gravitational constant — 45 kg * 9.8 m/s^2 = 441 N.

c. The weight component parallel to the plane can be found by multiplying the weight by sin(55 degrees) — 441 N * sin(55) = 360.56 N.

d. The normal force is calculated by taking the weight and multiplying by cos(55 degrees) — 441 N * cos(55) = 253.45 N.

e. To find the frictional force, we multiply the normal force by the frictional coefficient — 253.45 N * 0.24 = 60.83 N.

f. The acceleration of the crate can be found by subtracting the frictional force from the weight component along the plane, and then dividing by the mass of the crate — (360.56 N - 60.83N)/45 kg = 6.66 m/s^2.

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