A constant force acts on an object of mass 5kg for a duration of 2s. It increases the object’s velocity from 3 m/s to 7 m/s. Find the magnitude of applied force. Now, if the force was applied for a duration of 5s, what would be the final velocity of the object?
Answers
Answered by
7
Answer:
Given,
mass=5kg
t1 =2s
Initial velocity u=3m/s
Final velocity v=7m/s
t2 = 5s
So,
a = (v - u) / t = 7-3 / 2 = 2m/s²
So the magnitude of the applied force is 10N
And the final velocity after 5s is v
So,
v = u + at
v = 3+2 × 5
v = 13m/s
The final velocity after 5s is 13m/s
Hope its helpful ♥️♥️ Plzz follow me dear...☺️plzzz..
Answered by
0
Mass_ 5kg
Initial velocity (u)_3m/s
Final velocity (v)_7m/s
Time_ 2s
Acceleration(a)= v- u/t
=7-3/2
=2m/s^
Applied Force = Mass X Acceleration
= 5x2
= 10 N
Now, new time(t) =5s
Final velocity = u + at
= 3 + 2*5
= 3 + 10
= 13 m/s Ans
Similar questions