a constant force acts on an object of mass 10 kg for a duration of 4 second. it increases the object's velocity from 3 m/s to 7 m/s. calculate the applied force. now, if the force was applied for a duration of f second, then what would be the final velocity of the object?
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Answer:
509m/s
Explanation:
By work energy theorem:
4F= 1/2 * 10 ( 49 - 9)
= 1/2 * 400 = 200
F = 50 N
For F (50 s)
W= 50 * 50 = 2500
1/2 * 10 ( x - 9) = 2500
= > x - 9 = 500
= > x = 509m/s
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