Question

a constant force acts on an object of mass 5 kg for the duration of 2 sec. it increases the velocity from 3m/s to 7m/s. find the magnitude of the applied force.Now if the force were applied for duration of 5s what would we be the final velocity of the object?

Answer 1

☸Given:-
m=mass=5Kg
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as we know...

Force= Mass × Acceleration
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☸CASE 1:-
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Given:-
u=initial velocity=3m/s
v=final velocity=7m/s
t=time=2sec
a=acceleration=?

using formula:- v=u+at
=>7=3+a(2)
=>2a=4
=> a=2m/s^2
❣hence .... F= ma = 5×2=10N
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☸CASE 2:-
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given:-
same force for 5seconds
=> t=5s
=>a=2m/s^2
=>u=3m/s^2
=> v=???
using the same formula again... i.e.
=>v=u+at
=>v=3+2(5)
=>v=3+10
=>v=13m/s
❣hence .. the final velocity in 2nd part of the question will be 13m/s
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hope it helps you...
☺☺☺☺

Answer 2

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