Question

Easy velocity on inclined plane problem. A block of mass m slides down an inclined plane at an angle theta. The block is released at rest from a height h. Find the velocity, in components, of the block as it reaches the bottom of the inclined plane.

h=.5m

m=2.0kg

theta=30 degrees

Can you do this without using conservation of energy? That is the easiest way to do it, but I cannot find the solution using any other methods that I thought worked. I've tried solving for the time it takes for the block to fall a height h with a=gcos[theta], then plugging that into v=gtsin[theta]. I've also tried using v^2 = 2ah. I used both of those methods and tried to find the x and y components of the velocity. None of the components/magnitudes agree with eachother, nor do they agree with the conservation of energy technique.

My attempts:

Time to fall:

h=(1/2)a(t^2)

t=sqrt(2h/(gcos[30]) = .343s

Time to slide across:

Horizontal distance = d =hcot[30]

t=sqrt(2hcot[30]/(gcos[30]) = not .343

These two times should be the same, what have I done wrong?

Second attempt:

v^2 = 2ah

v=sqrt(2hgcos[30])=2.91m/s

which is almost the same as the velocity when using conservation of energy, so obviously that is wrong as well.

Answer 1

mgh = mv²/2

v = SQRT[2gh]

v =vcos[theta] i + vsin[theta] j