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shusha [124]
2 years ago
11

A 100 N force is applied to a 50 kg crate resting on a level floor. The coefficient of kinetic friction is 0.15. What is the acc

eleration of the crate after it begins to move
Physics
1 answer:
Nookie1986 [14]2 years ago
8 0

The acceleration of the crate after it begins to move is 0.5 m/s²

We'll begin by calculating the the frictional force

Mass (m) = 50 Kg

Coefficient of kinetic friction (μ) = 0.15

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

Normal reaction (N) = mg = 50 × 10 = 500 N

<h3>Frictional force (Fբ) =?</h3>

Fբ = μN

Fբ = 0.15 × 500

<h3>Fբ = 75 N</h3>

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

Frictional force (Fբ) = 75 N

Force (F) = 100 N

<h3>Net force (Fₙ) =?</h3>

Fₙ = F – Fբ

Fₙ = 100 – 75

<h3>Fₙ = 25 N</h3>

  • Finally, we shall determine the acceleration of the crate

Mass (m) = 50 Kg

Net force (Fₙ) = 25 N

<h3>Acceleration (a) =?</h3>

a = Fₙ / m

a = 25 / 50

<h3>a = 0.5 m/s²</h3>

Therefore, the acceleration of the crate is 0.5 m/s²

Learn more on friction: brainly.com/question/364384

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Answer and Explanation:

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As you may know the derivative of the position is the velocity and the derivative of the velocity is the acceleration. So we can get the velocity and the acceleration by deriving the position:

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Also, you may know these fundamental formulas:

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(a)

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(b)

v(9.7)=-(3\pi) (1.2) sin((3\pi *(9.7))+\frac{\pi}{5} ) \approx 9.1498 m/s

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(d)

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\phi = \frac{\pi}{5}

(e)

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(f)

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