Question

Can someone help me in this? mark as brainlist​

Answer 1

⭐ Newton's second law of motion :

Newton's second law of motion states that the net force acted on an object is directly proportional to the mass of the object multiplied by the acceleration.

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⭐ Given :

A constant that acts on an object of mass 5kg for duration of 4s increases the object's velocity from 6m/s to 14m/s.

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⭐ To find :

We shall have to find the magnitude of the force applied and the final velocity of the object of the force was applied for 10s.

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⭐ Solution :

According to the question, we have :

$\quad$ ☞ Mass of the object : m = 5kg

$\quad$ ☞ Time taken : t = 4s

$\quad$ ☞ Initial velocity : u = 6m/s

$\quad$ ☞ Final velocity : v = 14m/s

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⭐ Firstly, let's find the acceleration :

$\quad$ ☞ a = (v – u) / t

$\quad$ ☞ a = (14 – 6) / 4

$\quad$ ☞ a = 8 / 4

$\quad$ ☞ a = 2m/s²

$~$

⭐ Therefore, the acceleration of the object is 2m/s².

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⭐ Now, let's find the force acting on it :

$\quad$ ☞ F = m a

$\quad$ ☞ F = (5) (2)

$\quad$ ☞ F = 10N

$~$

⭐ Therefore, the force acting on the object is 10N.

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⭐ Also, we'll have to find the final velocity when the same force is acted on for 10s. So :

$\quad$ ☞ Mass : m = 5kg

$\quad$ ☞ Time : t = 10s

$\quad$ ☞ Force : f = 10N

$\quad$ ☞ Initial velocity : u = 6m/s

$\quad$ ☞ Final velocity : v

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⭐ We know that ‘a’ can also be written as (v – u) / t. Finding ‘v’ :

$\quad$ ☞ F = m × [ (v – u) / t ]

$\quad$ ☞ 10 = (5) × [ (v – 6) / 10 ]

$\quad$ ☞ 10 = (v – 6) / 2

$\quad$ ☞ 10 × 2 = v – 6

$\quad$ ☞ 20 + 6 = v

$\quad$ ☞ 26m/s = v

$~$

⭐ Therefore, the final velocity will be 26m/s when the force acts on for 10s.