A force acts for 1 s on a body of mass 2.0 kg initially at rest. The force is then withdrawn and the body attains a velocity of 2 m/s. Find the magnitude of the force.

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Solution

Here,
Acceleration, $a = \frac{v - u}{t}$
$= \frac{2 - 0}{1} = 2 m / s^{2}$
From the second law of motion, F = ma
= 2 $\times$ 2 = 4 N

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A force acts for 1 s on a body of mass 2.0 kg initially at rest. The force is then withdrawn and the body attains a velocity of 2 m/s. Find the magnitude of the force.

Here, Acceleration, a=v−ut =2−01=2 m/s2 From the second law of motion, F = ma = 2×2 = 4 N

Here,
Acceleration, $a = \frac{v - u}{t}$
$= \frac{2 - 0}{1} = 2 m / s^{2}$
From the second law of motion, F = ma
= 2 $\times$ 2 = 4 N