A 8 kg of ice, released from rest at the top of a 1.50 m long frictionless ramp, slides downhill, reaching speed of 2.50 m/s at the bottom. what is the angle between the ramp and the horizontal?
2007-03-22 22:32:58 · 3 answers · asked by biscuits 2 in Science & Mathematics ➔ Physics
lets t he time needed for the ice to arrive at the bottom
the ice is accelerated at an acceleration gsin â where â is the angle
so you have h = 1/2 (gsinâ) *t^2
you have also v = gsinâ *t or t = v/gsinâ
--->t^2 = v^2/(gsinâ)^2
h =v^2/2(gsinâ) giving sinâ = v^2/2gh
sinâ = 2.5^2/2*9.81*1.5 = 0.212
the angle is 12.3°
2007-03-22 22:57:26 · answer #1 · answered by maussy 7 · 0⤊ 0⤋
use conservation of energy to find vertical height
mgh=1/2mv^2
h=v^2/2g
sint=h/1.5 t is the angle
2007-03-23 05:47:12 · answer #2 · answered by tarundeep300 3 · 0⤊ 0⤋
LOL I was never any good at Physics sorry hun. :)
2007-03-23 05:43:06 · answer #3 · answered by flower19602003 5 · 0⤊ 0⤋