Question

A body of mass 10kg, moving with a velocity of 5 m/s comes to rest in 2 second. The
magnitude of force acting on the body is​

Answer 1

Answer:

25N

Explanation:

F=mà

F=10 ×5/2......

Answer 2

Required Solution:

Given:

• Mass of the body (m) = 10kg
• Initial velocity (u) = 5m/s
• Final velocity (v) = 0 [As it comes to rest. ]
• Time taken (t) = 2sec

To calculate:

• Force applied (F)

Calculation:

We know that,

• F = ma

Where,

• F = Force
• m = mass
• a = acceleration

Firstly we need to calculate acceleration:

By using the first equation of motion:

• v = u + at

→ 0 = 5 + ( a × 2 )

→ 0 = 5 + 2a

→ 2a = 0-5

→ 2a = -5

→ a = $\dfrac{-5}{2}$

Now we know that,

• F = ma

→ F = 10 × $\dfrac{-5}{2}$

→ F = 5 × (-5)

→ F = -25 N ⠀⠀⠀[Answer]

Therefore, force applied on the body is -25 N

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More information!

Equations of motion:

• v = u + at
• s = ut + ½at²
• v² – u² = 2as

Where,

• ★ v = final velocity = m/s
• ★ u = initial velocity = m/s
• ★ a = acceleration = m/s²
• ★ s = distance/displacement = m
• ★ t = time = sec

Remember that!

• When a body starts from rest, its initial velocity is 0.
• When a body comes to stop or applies breaks, its final velocity is 0.

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