Answer:
A: The acceleration is 7.7 m/s up the inclined plane.
B: It will take the block 0.36 seconds to move 0.5 meters up along the inclined plane
Explanation:
Let us work with variables and set
[tex]m=5kg\\\\F_H=100N\\\\\mu=0.3\\\\\theta=37^o.[/tex]
As shown in the attached free body diagram, we choose our coordinates such that the x-axis is parallel to the inclined plane and the y-axis is perpendicular. We do this because it greatly simplifies our calculations.
Part A:
From the free body diagram we see that the total force along the x-axis is:
[tex]F_{tot}=mg*sin(\theta)+F_s-F_Hcos(\theta).[/tex]
Now the force of friction is [tex]F_s=\mu*N,[/tex] where [tex]N[/tex] is the normal force and from the diagram it is [tex]F_y=mg*cos(\theta).[/tex]
Thus [tex]F_s=\mu*N=\mu*mg*cos(\theta).[/tex]
Therefore,
[tex]F_{tot}=mg*sin(\theta)+\mu*mg*cos(\theta)-F_Hcos(\theta)\\\\=mg(sin(\theta)+\mu*cos(\theta))-F_Hcos(\theta).[/tex]
Substituting the value for [tex]F_H,m,\mu, and \:\theta[/tex] we get:
[tex]F_{tot}= -38.63N.[/tex]
Now acceleration is simply
[tex]a=\frac{F_H}{m} =\frac{-38.63N}{5kg} =-7.7m/s.[/tex]
The negative sign indicates that the acceleration is directed up the incline.
Part B:
[tex]d=\frac{1}{2} at^2[/tex]
Which can be rearranged to solve for t:
[tex]t=\sqrt{\frac{2d}{a} }[/tex]
Substitute the value of [tex] d=0.50m[/tex] and [tex] a=7.7m/s[/tex] and we get:
[tex]t=0.36s.[/tex]
which is our answer.
Notice that in using the formula to calculate time we used the positive value of [tex]a[/tex], because for this formula absolute value is needed.
