V^2 = 2gd = 2 * 9.8 * 2 = 39.2m.
V = sqrt(39.2) = 6.26m/s.
a 5kg suitcase slides 2m down a frictionless ramp incline at 45 degress. How fast is it going at the bottom?
2 answers
2m*sin(45)= length of fall (LOF)
this needs to be plugged into equation henry posted. The suitcase does not fall 2m unimpeded. It slides down the ramp at 45 degrees. Thus trig is needed to solve.
this needs to be plugged into equation henry posted. The suitcase does not fall 2m unimpeded. It slides down the ramp at 45 degrees. Thus trig is needed to solve.
1 answer