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The coefficient of kinetic friction is 0.21, for a 8 kg block on the floor. It is being pulled by a 17 N force, what is the acceleration of the box?

Answer :

EDUARD22SLYGO

The acceleration of the box is 0.025 m/s²

We'll begin by calculating the the frictional force

Coefficient of kinetic friction (μ) = 0.21

Mass (m) = 8 Kg

Acceleration due to gravity (g) = 10 m/s²

Normal reaction (N) = mg = 8 × 10 = 80 N

Frictional force (Fբ) =?

Fբ = μN

Fբ = 0.21 × 80

Fբ = 16.8 N

  • Next, we shall determine the net force acting on the box

Frictional force (Fբ) = 16.8 N

Force (F) = 17 N

Net force (Fₙ) =?

Fₙ = F – Fբ

Fₙ = 17 – 16.8

Fₙ = 0.2 N

  • Finally, we shall determine the acceleration of the box

Mass (m) = 8 Kg

Net force (Fₙ) = 0.2 N

Acceleration (a) =?

a = Fₙ / m

a = 0.2 / 8

a = 0.025 m/s²

Therefore, the acceleration of the box is 0.025 m/s²

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