Question

A constant force acts on an
object of mass 5 kg for a duration 25.It increases the objects velocity
from 3ms to 7ms-1.Find the
magnitude of force​

Answer 1

Mass of the object (m) = 5 kg

Time duration (t) = 25 s

Initial velocity (u) = 3 m/s

Final velocity (v) = 7 m/s

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Force (F) : It is defined as product of mass and acceleration

➠ F = m × a

Acceleration (a) : It is defined as rate of change in velocity

➠ a = ( v - u ) / t

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$:\implies \sf F=m\times a$

$:\implies \sf F=m\times \bigg(\dfrac{v-u}{t}\bigg)$

$:\implies \sf F=5 \times \bigg( \dfrac{7-3}{25}\bigg)$

$:\implies \sf F= \dfrac{4}{5}$

$:\implies \bf \orange{F=0.8\ N}\ \; \bigstar$

So , Magnitude of the force is 0.8 N .

Answer 2

Given :

• A constant force acts on an object of mass 5 kg for a duration 25
• It increases the objects velocity from 3 ms to 7 ms^{-1}

To Find :

• The magnitude of force

Solution :

We have,

• Mass of the object (m) = 5 kg
• Time taken (t) = 25 s
• Initial velocity (u) = 3 m/s
• Final velocity (v) = 7 m/s

We know that,

$\boxed{\tt F = m\bigg(\dfrac{v - u}{t}\bigg)}$

Substituting the given values we get,

$:\implies\sf F = 5\bigg(\dfrac{7 - 3}{25}\bigg)$

$:\implies\sf F = 5\bigg(\dfrac{4}{25}\bigg)$

$:\implies\sf F = \cancel5 \times \dfrac{4}{\cancel{25}}$

$:\implies\sf F = \dfrac{4}{5}$

$:\implies\sf F = 0.8 \: N$

Hence, the magnitude of force is 0.8 N