1. Apply a constant force of 50 N directed to the right of the 50 kg Box. (2 pts)

Physics

1 answer:

kotegsom [21]2 years ago

7 0

As the box is moving with a constant velocity, the two forces acting on the box are canceling each other.

Then friction force = 80 Newtons but in the opposite direction.

Friction force = Mu * Normal force exerted by ground = Mu * weight of box

So we find Mu.

Mu = coefficient of friction between box and horizontal surface

= Force of friction / weight = 80 / 50 * 9.81 = 0.163

When an identical box is placed on top, the force of friction is

= Mu * total weight = 0.163 * (50+50) * 9.81 = 159.9 Newtons

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3 years ago

On a horizontal surface is located

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By Newton's second law, the net vertical force acting on the object is 0, so that

n - w = 0

where n = magnitude of the normal force of the surface pushing up on the object, and w = weight of the object. Hence n = w = mg = 196 N, where m = 20 kg and g = 9.80 m/s².

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80 N = µ (196 N) → µ = (80 N)/(196 N) ≈ 0.408

Moving at constant speed, there is a kinetic friction force of 40 N opposing the object's motion, so that the coefficient of kinetic friction ν is

40 N = ν (196 N) → ν = (40 N)/(196 N) ≈ 0.204

And so the closest answer is C.

(Note: µ and ν are the Greek letters mu and nu)

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3 years ago

Help me fast and i brainliest.

nexus9112 [7]

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