Question

A block of ice with mass 2.00 kg slides 0.830 m down an inclined plane that slopes downward at an angle of 31.0 ∘ below the horizontal.

If the block of ice starts from rest, what is its final speed? You can ignore friction.

Answer 1

The part of the block’s weight that is forcing the block to accelerate down the 36.9º angle inclined plane = mass * 9.8 * sin 31º

This force, parallel to the surface of the inclined plane = 2 * 9.8 * sin 31º

This force causes the 2 kg block to accelerate down the inclined plane!

Force = mass * acceleration
2 * 9.8 * sin 31º = 2 * a
a = 9.8 * sin 31º

Distance = ½ * a * t^2
0.830 = ½ * 9.8 * sin 31º * t^2

t = (2*0.830)/(9.8*0.5150)^(1/2)

t = 0.5735 seconds

final velocity = a * t
final velocity = 9.8 * sin 31º * 0.5735 s = 2.89 m/s

if you wanted to use conservation of energy:

PE = mass * 9.8 * height
KE = ½ * mass * velocity^2
PE = KE
mass * 9.8 * height = ½ * mass * velocity^2
mass cancels

9.8 * height = ½ * velocity^2
19.6 * height = velocity^2

the height of the inclined plane = 0.830 * sin 31º
19.6 * (0.830 * sin 31º) = velocity^2

v = 2.89 m/s