Starting from rest, a 11.70 kg suitcase slides 3.14 m down a frictionless ramp inclined at 40° from the floor. The suitcase then slides an additional 4.68 m along the floor before coming to a stop.
(a) Determine the suitcase's speed at the bottom of the ramp.
(b) Determine the coefficient of kinetic friction between the suitcase and the floor.
(c) Determine the change in mechanical energy due to friction.
1 answer
1)v=sqrt(2*9.8*3.14sin40)=6.29m/s (2)umgd=1/2mv^2 u*9.8*4.68=0.5(6.29)^2 u=0.43 (3)change in mechanical energy=workdone by friction=0.43*11.7*9.8*46.8=2307.4J
2 answers