A block of ice with mass 2.00kg slides 0.830 m down an inclined plane that slopes downward at an angle of 30.8 degrees below the horizontal.
If the block of ice starts from rest, what is its final speed? You can ignore friction.

Question

A block of ice with mass 2.00kg  slides 0.830 m down an inclined plane that slopes downward at an angle of 30.8 degrees below the horizontal.
If the block of ice starts from rest, what is its final speed? You can ignore friction.

anonymous · Answer

If you draw an energy flow diagram, you can see that the potential energy at the top of the plane is equal to the kinetic energy at the bottom.

So Ug=K.

mgh=1/2 mv^2

cross out mass on both sides to get
gh=1/2 v^2

arrange the equation to get 
$$v=\sqrt{2gh}$$

for h, use 
sin 30.8=y/.83

then you'll get .62 m as the height. 

now plug everything into the equation to get v=3.49m/s

anonymous · Answer

i got it like this mgdsin30.8-1/2mvsquared solving for vsquared i got 2.89m/s which is the correct answer, thank you though!

anonymous · Answer

ohhh i got confused with the position of the angle. okay i got the same answer after i recalculated the height. haha

anonymous · Answer

:)