A concrete slab of mass 400 kg accelerates down a concrete slope inclined at 35°. The u kinetic bn the slab and slope is 0.60. Determine the acceleration of the block.

Answer 1

I do not care how massive the block is.
I do care the g = about 9.81 m/s^2
Force down slope = m g sin 35
normal force = m g cos 35
friction force up slope = 0.60 m g cos 35
so
m g ( sin 35 - 0.60 cos 35) = m a
a = g (sin 35 - 0.60 cos 35)

Answer 2

M*g = 400 * 9.8 = 3920 N. = Wt. of slab,
Fp = 3920*sin35 = 2248.4 N. = Force parallel to incline,
Fn = 3920*Cos35 = 3211 N. = Normal force,
Fk = u*Fn = 0.6 * 3211 = 1927 N. = Force of kinetic friction,
Fp-Fk = M*a,
2248.4 - 1927 = 400*a.

Answer 3

ROFL :)

Answer 4

determine the acceleration of the block.

Answer 5

a=g(sin 35 -0.060 cos 35) =0.829

Answer 6

Acceleration 400kg 35%