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A 230 kg steel crate is being pushed along a cement floor. The force of friction is 480 N to the left and the applied force is 1869 N to the right. What is the acceleration of the crate?
a=(6.0 or 6.00 or 8.10 or 8.1)=m/s^2

Respuesta :

Eduard22sly

The acceleration of the steel crate, given the data from the question is 6.0 m/s²

How to determine the net force

  • Force to the left (Fբ) = 480 N
  • Force to the right (Fᵣ) = 1869 N
  • Net force (Fₙ) = ?

Fₙ = Fᵣ - Fբ

Fₙ = 1869 - 480

Fₙ = 1389 N

How to determine the acceleration

  • Mass (m) = 230 Kg
  • Net force (Fₙ) = 1389 N
  • Acceleration (a) = ?

Fₙ = ma

Divide both sides by m

a = Fₙ / m

a = 1389 / 230

a = 6.0 m/s²

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