A constant force acts on an object of mass 10 kg for a duration of 4 s. It increases the object’s velocity from 6 m s –1 to 14 m s -1 . Find the magnitude of the applied force. Now, if the force was applied for duration of 10 s, what would be the final velocity of the object?
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Answer:
F=20 N
v=34 m/s
Explanation:
a=(v-u)/t
=(14-6)/4
=8/4
=2 m/s^2
F=ma
=10 kg× 2m/s^2
=20 N
Now acceleration will be 2 m/s^2 if same fore is applied for 10 s.
a=(v-u)/t
2=(v-14)/10
20=v-14
v=20+14
v=34 m/s
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