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Vladimir79 [104]
3 years ago
12

Please help.: Sliding from left to right in a straight line on a horizontal steel surface, an aluminum block weighing 20 newtons

is acted on by a 2.4 newton friction force. The block will be brought to rest by the friction force in a distance of 10 meters. Determine the magnitude of the acceleration of the block as it is brought to rest by the friction force. (Show all work)
Physics
2 answers:
Vlad [161]3 years ago
7 0
Mass of the block = 2/g = 2/9.8 =0.204 kg. 

Frictional force = mass * acceleration 

a = F/ m where F is the frictional force = 2.4 

a = 2.4/ 0.204 = 11.76 m/s^2 
-BARSIC- [3]3 years ago
7 0

Answer:

-1.176 m/s²

Explanation:

Given:

weight of aluminium block is w = 20.0 N

frictional force, f = 2.4 N

final velocity, v = 0

distance covered, d = 10 m

From Newton's second equation of motion,

F = ma

where, m is the mass and a is the acceleration.

mass of the block, m g = 20.0 N

m = 2.04 kg

The block is brought to rest by friction force:

f =- m a

⇒ 2.4 N = -(2.04 kg)(a)

⇒a = -1.176 m/s²

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          cos 45 = F₂₄ₓ / F₂₄

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          Fₓ = -F₂₁₁ + F₂₄ cos 45

Y axis  

         F_y = - F₂₃ + F_{24y}

         F_y = -F₂₃ + F₂₄ sin 45

They indicate that the magnitude of all charges is the same, therefore

         F₂₁ = F₂₃

Let's use Coulomb's law

         F₂₁ = k q₁ q₂ / r₁₂²

       

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         r = a

         F₂₁ = k q² / a²

we calculate F₂₄

           F₂₄ = k q₂ q₄ / r₂₄²

the distance is

           r² = a² + a²

           r² = 2 a²

         

we substitute

           F₂₄ = k  q² / 2 a²

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          Fx = k \frac{q^2}{a^2}  ( -1 + ½ cos 45)

          F_y = k \frac{q^2}{a^2} ( -1 +  ½ sin 45)    

         

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            Fₓ = 8.40 10¹¹ (½ 0.707 - 1)

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remember cos 45 = sin 45

             F_y = - 5.43 10¹¹  N

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a) F = Fₓ î + F_y ^j

          F = -5.43 10¹¹ (i + j)   N

b) In the form of module and angle.

For the module we use the Pythagorean theorem

          F = \sqrt{F_x^2 + F_y^2}

          F = 5.43 10¹¹  √2

          F = 7.68 10¹¹ N

in angle is

           θ = 45º

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2 years ago
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