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

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

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explanation: