Question

4. A 20-kg box sits on an incline of 30° from the horizontal. The coefficient of friction between the box and the incline is o.30. Find the acceleration of the box down the incline.

Answer 1

The box accelerates down the incline at 2.4m/s^2.

How do you find the acceleration of an incline box?

A mass m particle will slide down a smooth inclined plane if it is released onto it with a friction force F=0. We resolve in the direction of motion to determine the particle's acceleration as it slides. F = ma, mg cos(90) = ma, gg cos(90) = a, gg sin() = a.

By using zigmaFx = max, we will determine the acceleration.

However, we must first determine the friction force Ff.

Because cos 30 degree = 0.866 and Fy = ma y = 0 result in FN - 0.87mg = 0, we may calculate FN as (0.87)(20Kg)(9.81 m/s2) = 171N.

From Ff = mue FN = 0.30)(171 N)= 51N, we can now calculate Ff.

We get Ff - 0.50mg = ma x 51N - (0.50)(20)(9.81)N = (20kg)(ax) from the expression zigmaFx = max, where ax = -2.35 m/s2.

At 2.4 m/s2, the box quickens its descent of the hill.

To learn more about acceleration refer to:

brainly.com/question/460763

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