normal force = 400 g cos 35
friction force up slope = .6 (400 g) cos 35
weight component down slope = 400 g sin 35
400 a = 400 g sin 35 - .6 (400 g cos 35)
a = g (sin 35 - .6 cos 35) = .082 g
a concrete slab of mass 400 kg accelerates down a concrete slope inclined at 35 degrees. the coefficient of kinetic friction between the slab and slope is 0.60. determine the acceleration of the block
10 answers
the momentum after the collision
400kg
I do not get it ! Would you say it works again?
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What is this thing
I don't understand clearly
Mohammed
6 answers