A force acts for 0.1 sec on a body of mass 2kg initially at rest. The force is then withdrawn and the body moves with a velocity of 2m/s . Find the magnitude of force

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Solution

We have,

Mass, m = 2 kg

Initial velocity, u = 0

Final velocity, v = 2 m/s

Time, t = 0.1 s

Now, Acceleration, a = (v - u)/t

=> a = (2 - 0)/0.1 = 20 m/s²

So, force, F = ma = (2)(20) = 40 N

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A force acts for 0.1 sec on a body of mass 2kg initially at rest. The force is then withdrawn and the body moves with a velocity of 2m/s . Find the magnitude of force

We have, Mass, m = 2 kg Initial velocity, u = 0 Final velocity, v = 2 m/s Time, t = 0.1 s Now, Acceleration, a = (v - u)/t => a = (2 - 0)/0.1 = 20 m/s² So, force, F = ma = (2)(20) = 40 N

We have,

Mass, m = 2 kg

Initial velocity, u = 0

Final velocity, v = 2 m/s

Time, t = 0.1 s

Now, Acceleration, a = (v - u)/t

=> a = (2 - 0)/0.1 = 20 m/s²

So, force, F = ma = (2)(20) = 40 N