V^2 = Vo^2 + 2g*h = 0 + 19.6*0.65 = 12.74
V = 3.57 m/s.
A small block of mass .450 kg is sliding down a ramp inclined at an angle of 53 degrees to the horizontal. it starts to slide from a vertical height of .65 meters. Find the Velocity when it slides off the ramp.
The way I approached this problem was by sing Vf^2-VI^2=2ad. Because velocity initial is 0 its vf^2=2ad. I plugged in vf^2=(2)(9.8)(sin53)(.45)(.81). I got .81 meters by measuring the distance the block traveled downwards by using sin53=.65/x. I ended up with the answer 2.4 m/s. Is this correct?
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