A constant force acts on an object of mass 5kg for a duration of 2 sec. it increases the object's velocity from 3m/sec to 7m/sec.find the magnitude of applied force.now if the force were applied for a duration 5sed what will be its final velocity?
Answers
Answered by
6
Let the force be F
m=5kg
t=2s
velocity is increasing from 3m/s to 7m/s
a=(7-3)/2
⇒4/2
⇒2m/s²
F=mxa
⇒5x2
10N
Now, force is applied for 5 sec, so final velocity after 5sec is
V=u+at
⇒3+2(5)
⇒13m/s
m=5kg
t=2s
velocity is increasing from 3m/s to 7m/s
a=(7-3)/2
⇒4/2
⇒2m/s²
F=mxa
⇒5x2
10N
Now, force is applied for 5 sec, so final velocity after 5sec is
V=u+at
⇒3+2(5)
⇒13m/s
Similar questions