Question

A block of ice with mass 2.00 kg slides 0.750 m down an inclined plane that slopes downward at an angle of 36.9 degrees below the horizontal. If the block of ice starts from rest, what is its final speed? You can ignore friction.

Answer 1

Answer: $V_{f}=2.96m/s$    

Firstly we have to draw the Free Body Diagram (FBD) as shown in the figure attached.

Where the weight $w$ of the block has an x-component and y-component:

$w_{x}=wsin(\theta)$    (1)

$w_{y}=wcos(\theta)$    (2)

As well as the Normal Force $N$:

$N_{x}=Nsin(\theta)$    (3)

$N_{y}=Ncos(\theta)$    (4)

In addition, we know $N=w$, then $\sum F_{y}=0$

In the X-component:

$\sum F_{x}=m.a$

$m.a=w_{x}$    (5)

Substituting (1) in (5):

$wsin(\theta)=m.a$    (6)

In addition, we know $w=m.g$, where $m$ is the mass of the block and $g$ the gravity acceleration, which is equal to $9.8m/{s}^{2}$  

So:

$m.g.sin(\theta)=m.a$   (7)

$a=g.sin(\theta)$    (8)

$a=5.88m/{s}^{2}$    (9)   >>>>This is the acceleration of the block

On the other hand, we have the following equation that expresses a relation between the distance $d$ with the acceleration $a$ and time $t$:

$d=\frac{1}{2}a{t}^{2}$   (10)

We already know the value of  $d$ and calculated $a$, we have to find $t$:

$t=\sqrt{\frac{2d}{a}}$   (11)

$t=\sqrt{\frac{2(0.75m)}{5.88m/{s}^{2}}}$   (12)

$t=0.50s$   (13) >>>This is the time it takes to the block to go from the initial velocity $V_{o}$ to its final velocity $V_{f}$

If the acceleration is the variation of the velocity in time, we can use the following equation to find $V_{f}$:

$V_{f}-V_{o}=a.t$   (13)

If $V_{o}=0$

$V_{f}=a.t$   (14)

$V_{f}=(5.88m/{s}^{2})(0.50s)$   (15)

Finally we get the value of the Final Velocity of the block:

$V_{f}=2.96m/s$