Question

A body of mass 20×10^-3 when acted upon by a force for 4 second, attains a velocity of 100m/s .if the same force applied for 2 minutes on a body of mass 10 kilogram at rest what will be its velocity

Answer 1

Solution :

$\dag$Case I:

As per the given data,

• Mass (m)= 0.020 kg
• Time(t) = 4 s
• Final velocity (t)= 100 m/s
• Initial velocity (u)= 0 m/s

First, we need to find the acceleration of the body.

By using first equation of motion,

• v = u + at

On rearranging,

⇒ a = v - u / t

Now let's substitute the given values,

⇒ a = 100 - 0 / 4

⇒ a = 25 m/s²

Now,

By applying Newton's second law,

• F = ma

⇒ F = 0.020 x 25

⇒ F = 0.5 N

$\dag$Case II:

Given data,

• Force(F) = 0.5 N
• Time (t) = 2 min = 2 x 60 = 120 s
• Mass(M) = 10 kg
• Initial velocity = 0 m/s [rest]

By applying the first equation of motion,

• v = u + at

On rearranging,

⇒ a = v - u / t

⇒ a = v / t

Now,

By using Newton's second law we get,

⇒ F = Ma

⇒ 0.5 = 10 X v / 120

⇒ v = 0.5 x 12

⇒ v = 6 m /s

Answer 2

Explanation:

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Case I:

As per the given data,

Mass (m)= 0.020 kg

Time(t) = 4 s

Final velocity (t)= 100 m/s

Initial velocity (u)= 0 m/s

First, we need to find the acceleration of the body.

By using first equation of motion,

v = u + at

On rearranging,

⇒ a = v - u / t

Now let's substitute the given values,

⇒ a = 100 - 0 / 4

⇒ a = 25 m/s²

Now,

By applying Newton's second law,

F = ma

⇒ F = 0.020 x 25

⇒ F = 0.5 N

\dag† Case II:

Given data,

Force(F) = 0.5 N

Time (t) = 2 min = 2 x 60 = 120 s

Mass(M) = 10 kg

Initial velocity = 0 m/s [rest]

By applying the first equation of motion,

v = u + at

On rearranging,

⇒ a = v - u / t

⇒ a = v / t

Now,

By using Newton's second law we get,

⇒ F = Ma

⇒ 0.5 = 10 X v / 120

⇒ v = 0.5 x 12

⇒ v = 6 m /s