Question

A box of wood is kept on a 30 degree slope. If the coefficients of friction is 0.1, What is the downward acceleration of the wooden box?

Answer 1

heya yor answer is__________
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Given, μ = 0.1
θ = 300
Downward acceleration of the wooden box is calculated by the formula :
a = g ( sin ​θ - ​μ cos ​​θ )
Put given values in the given formula to get your answer.

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Answer 2

Answer:

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Explanation:

The forces acting on the box are:

(1) weight of the box vertically down ward= mg

(2) force of normal reaction by the slope=R.

(3) the frictional force upward parallel to the inclined surface=f

We resolve the weight into two components: mg sin 30° , parallel downward to the inclined plane. The second component is mg cos30° . This component is perpendicular to the inclined plane and it balances the normal force, R. So,

R=mg cos30° and hence frictional force f=uR=umgcos30°……………(1)

Now, we can write the eq. of motion as follows:

ma=mg sin30° - u mgcos 30°, or

a=9.8/2-(0.1)(9.8)(sqrt3/2). You may calculate a from this equation.