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A 70 kg box is slid along the floor by a 400 N force. The coefficient of friction between the box and the floor is 0.50 when the box is sliding. Find the acceleration of the box

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PastaMario
We have to employ Newton's Second Law and Third Law here: force = mass * acceleration, and every force produces an equal and opposite force.

—The forces vertically are EQUAL, so no acceleration is occurring in this direction. This also means that the force of gravity and the normal force must be equal in magnitude.
—The forces horizontally are NOT equal, so there will be acceleration on this box in this direction. The applied force is going to be larger than the friction force, and the friction force opposes the direction of the applied force.

First, we find the weight of this box, which is given by mass * acceleration due to gravity. On Earth, acceleration due to gravity is about 9.8 m/s^2.
So, 70 kg * 9.8 m/s^2 = 686 N. This force points downwards.
The normal force is equal in magnitude to this, so it is also = 686 N, but points upwards.

The force of friction is given by force(friction) = µ * force(normal). In this problem, µ = 0.50. Then, we plug in to get 0.50 * 686 N = 343 N. This is the force of friction that opposes the motion of the box.

And finally, we are given the applied force, which is 400 N.
Now we make a Newton's second law equation.

force(applied) - force(friction) = mass * acceleration

400 N - 343 N = 70 kg * a

a = .8142 m/s^2

I really hope this is the right answer; it's 4 AM for me lol
Eduard22sly

The acceleration of the box of mass 70 Kg sliding along the floor and havin 0.5 coefficient of friction is 0.81 m/s².

  • We'll begin by calculating the frictional force opposing the motion. This can be obtained as follow:

Mass (m) = 70 Kg

Coefficient of friction (µ) = 0.5

Acceleration due to gravity (g) = 9.8 m/s²

Normal reaction (N) = mg = 70 × 9.8 = 686 N

Frictional force (F) =?

F = µN

F = 0.5 × 686

F = 343 N

Thus the frictional force is 343 N.

  • Next, we shall determine the net force acting on the box. This can be obtained as follow:

Force applied (F) = 400 N

Frictional force (Fբ) = 343 N

Net force (Fₙ) =?

Fₙ = F – Fբ

Fₙ = 400 – 343

Fₙ = 57 N

Thus, the net force acting on the box is 57 N

  • Finally, we shall determine the acceleration of the box. This can be obtained as follow:

Mass (m) = 70 Kg

Net force (Fₙ) = 57 N

Acceleration (a) =?

Fₙ = ma

57 = 70 × a

Divide both side by 70

a = 57 / 70

a = 0.81 m/s²

Therefore, the acceleration of the box is 0.81 m/s².

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