Answer :
Answer: [tex]V_{f}=2.96m/s[/tex]
Firstly we have to draw the Free Body Diagram (FBD) as shown in the figure attached.
Where the weight [tex]w[/tex] of the block has an x-component and y-component:
[tex]w_{x}=wsin(\theta)[/tex] (1)
[tex]w_{y}=wcos(\theta)[/tex] (2)
As well as the Normal Force [tex]N[/tex]:
[tex]N_{x}=Nsin(\theta)[/tex] (3)
[tex]N_{y}=Ncos(\theta)[/tex] (4)
In addition, we know [tex]N=w[/tex], then [tex]\sum F_{y}=0[/tex]
In the X-component:
[tex]\sum F_{x}=m.a[/tex]
[tex]m.a=w_{x}[/tex] (5)
Substituting (1) in (5):
[tex]wsin(\theta)=m.a[/tex] (6)
In addition, we know [tex]w=m.g[/tex], where [tex]m[/tex] is the mass of the block and [tex]g[/tex] the gravity acceleration, which is equal to [tex]9.8m/{s}^{2}[/tex]
So:
[tex]m.g.sin(\theta)=m.a[/tex] (7)
[tex]a=g.sin(\theta)[/tex] (8)
[tex]a=5.88m/{s}^{2}[/tex] (9) >>>>This is the acceleration of the block
On the other hand, we have the following equation that expresses a relation between the distance [tex]d[/tex] with the acceleration [tex]a[/tex] and time [tex]t[/tex]:
[tex]d=\frac{1}{2}a{t}^{2}[/tex] (10)
We already know the value of [tex]d[/tex] and calculated [tex]a[/tex], we have to find [tex]t[/tex]:
[tex]t=\sqrt{\frac{2d}{a}}[/tex] (11)
[tex]t=\sqrt{\frac{2(0.75m)}{5.88m/{s}^{2}}}[/tex] (12)
[tex]t=0.50s[/tex] (13) >>>This is the time it takes to the block to go from the initial velocity [tex]V_{o}[/tex] to its final velocity [tex]V_{f}[/tex]
If the acceleration is the variation of the velocity in time, we can use the following equation to find [tex]V_{f}[/tex]:
[tex]V_{f}-V_{o}=a.t[/tex] (13)
If [tex]V_{o}=0[/tex]
[tex]V_{f}=a.t[/tex] (14)
[tex]V_{f}=(5.88m/{s}^{2})(0.50s)[/tex] (15)
Finally we get the value of the Final Velocity of the block:
[tex]V_{f}=2.96m/s[/tex]
