Asked by Ashley
A 400 gram block slides down a 30 degree incline from rest and reaches the bottom with a speed of 4.50 m/s. If the height through which the block falls is 2.0 meters, find the average frictional force that acted on the block.
Answers
Answered by
Henry
L = h/sin A = 2/sin 30 = 4 m. = Length of incline.
KE = mg*h-Fk*L
0.5m*V^2 = mg*h-Fk*l
0.2*4.50^2 = 3.92*2 - Fk*4
4.05 = 7.84 - 4Fk
4Fk = 7.84-4.05 = 3.79
Fk = 0.948 N. = Force of kinetic friction.
KE = mg*h-Fk*L
0.5m*V^2 = mg*h-Fk*l
0.2*4.50^2 = 3.92*2 - Fk*4
4.05 = 7.84 - 4Fk
4Fk = 7.84-4.05 = 3.79
Fk = 0.948 N. = Force of kinetic friction.
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