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A 18.0 kg block slides down an inclined plane (angle of inclination is 36 degrees). If the coefficient of kinematic friction is 0.55, what is the acceleration of the block? Consider g = 9.8 m/s². Express your answer correct to 2 decimal places. Add the appropriate unit(s).​

User Elia Ahadi
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Answer:

Step-by-step explanation:

The force of gravity acting on the block can be resolved into two components, one parallel to the incline and one perpendicular to the incline. The perpendicular component is balanced by the normal force of the incline, and the parallel component is opposed by the force of friction. The force of friction is given by:

F_friction = coefficient_of_friction * F_norm

where F_norm is the normal force of the incline. The normal force is equal in magnitude and opposite in direction to the perpendicular component of the force of gravity, which is:

F_perpendicular = m * g * cos(theta)

where m is the mass of the block, g is the acceleration due to gravity, and theta is the angle of inclination.

The parallel component of the force of gravity is:

F_parallel = m * g * sin(theta)

The net force acting on the block is:

F_net = F_parallel - F_friction

Using Newton's second law, F = m * a, we can solve for the acceleration of the block:

a = F_net / m

Substituting the expressions for F_parallel and F_friction, we get:

a = [m * g * sin(theta) - coefficient_of_friction * m * g * cos(theta)] / m

Simplifying, we get:

a = g * [sin(theta) - coefficient_of_friction * cos(theta)]

Substituting the given values, we get:

a = 9.8 m/s^2 * [sin(36°) - 0.55 * cos(36°)] = 6.43 m/s^2

Therefore, the acceleration of the block is 6.43 m/s^2.

User Duc Manh Nguyen
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