Question

A constant force act on an object of mass 5 kg for a time of 2 seconds It increases the object's velocity from 4 m/s to 10 m/s. Find the magnitude of the applied force

Answer 1

Answer:

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

Answer 2

Answer:

10 N

Explanation:

Given,

mass=5kg

t1  =2s

Initial velocity u=3m/s

Final velocity v=7m/s

t2  =5s

So,

Let the Force be F

Let the acceleration be a

So,

a=  t(v−u)  =  2

(7−3)  =2m/s  

So the magnitude of the applied force is 10N

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