Question

a boy weighing 100kg moves at a speed of 10m/s and stops at adistance of 100m with deaccelating slowly uniformly what is the force applied
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Answer 1

Answer:

Answer is 50 N of the question

Answer 2

Mass of boy = 100 kg

Initial velocity = 10 m/s

Final velocity = 0 m/s (rest)

Distance travelled = 100 m

We know,

2a∆x = v(f)² - v(i)²

⇒ a = - v(i)²/2∆x

⇒ a = - (10 m/s)²/(2 × 100 m)

⇒ a = - (100 m²/s²)/(200 m)

⇒ a = - ½ m/s²

⇒ a = - 0.5 m/s².

Now we know, F = ma

⇒ F = (100 kg)(- 0.5 m/s²)

⇒ F = - 50 N.

∴ He exerted a force of 50 N in the opposite direction of it's motion.