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

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

Where is full answer?

I do not get it ! Would you say it works again?

Really it's not clear

400kg

I don't understand clearly

What is this thing

the momentum after the collision

Mohammed

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