A3000 kg truck is sliding down a hill that is at an angle of 40 degrees above the horizontal. if the coefficient of friction between the truck and the hill is 0.2, what is the acceleration of the truck? ignore all other forms of friction. show all work

4.8 m/s²

Explanation:

Draw a free body diagram of the truck.  There are three forces:

Weight force mg pulling down

Normal force Fn pushing perpendicular to the hill

Friction Force Fn μ pushing up the hill

Sum of the forces in the perpendicular direction:

∑F = ma

Fn − mg cos θ = 0

Fn = mg cos θ

Sum of the forces in the parallel direction (taking down the hill to be positive):

∑F = ma

mg sin θ − Fn μ = ma

mg sin θ − mg μ cos θ = ma

g sin θ − g μ cos θ = a

a = g (sin θ − μ cos θ)

Substitute:

a = (9.8) (sin 40° − 0.2 cos 40°)

a = 4.8 m/s²

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6/12=1/2

step-by-step explanation: