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A 6.0-kg block initially at rest is pulled to the right along a frictionless, horizontal surface by a constant horizontal force of magnitude 12 N. Find the block’s speed after it is has moved through a horizontal distance of 3.0 m

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Kazeemsodikisola

Answer:

Explanation:

Given that,

Block of mass M= 6kg

Initially at rest u =0

A force F = 12N is used in pulling it along the horizontal

Distance moved s = 3m

Speed.

Using Newton second law

ΣFx = ma

12 = 6a

Then, a = 12/6

a = 2m/s²

Then, using equation of motion

v² = u² + 2as

v² = 0² + 2×2×3

v² = 0 + 12

v = √ 12

v = 3.46 m/s

Cricetus

The block's speed after it moved will be "3.46 m/s".

Newton second law

Block's mass, M = 6 kg

Force, F = 12 N

Distance moved, s = 3 m

Initially at rest, u = 0

By using Newton second law, we get

→ ∑Fx = ma

By substituting the values,

     12 = 6a

       a = [tex]\frac{12}{6}[/tex]

          = 2 m/s²

hence,

By using Equation of motion,

→ v² = u² + 2as

By putting the values,

       = (0)² + 2 × 2 × 3

       = 0 + 12

    v = √12

       = 3.46 m/s

Thus the response above is correct.  

Find out more information about Equation of motion here:

https://brainly.com/question/25951773

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