# A constant force acts on an object of mass 5 kg for a duration of 2 seconds. It increases the object's velocity from 3 m/s to 7 m/s. Find the magnitude of the applied force. Now if the force were applied for a duration of 5 s what would be the final velocity of the object????????

2
by chandraharish

2014-10-26T13:08:11+05:30

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Let the Force be F
mass = 5 kg
time(t) = 2 s
initial velocity(u) = 3 m/s
final velocity(v) = 7 m/s
so let the acceleration be a
so a = (v - u)/t = (7 - 3)/2 = 2 m/s²

So the magnitude of the applied force is 10 N

and the final velocity after 5 s is v

so v = u + at
v = 3 + 2 x 5
v = 13 m/s

The final velocity after 5 s is 13 m/s
2014-10-26T13:10:22+05:30

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Mass= 5 kg
u = 3m/s
v = 7 m/s
t = 2s

acceleration = (v-u)/t = (7-3)/2 = 4/2 = 2m/s²
force = ma = 5×2 = 10N

now if the force had been applied for 5 seconds
final velocity = 3 + 2×5 = 13 m/s
you missed
ok
in writting the initial velocity
the word "NOW" if the force applied for 5 sec, does it mean the acceleration is given for 7 seconds ? perhaps ?
well, if this is a problem from the book, the answer given there depends on what the author thinks..