Question

A constant force acts ona an object of mass 5Kg for a duration of 2 sec. It increases the object velocity from 3ms-1 to 7ms-1. find the magnitude of the applied force. Now if the force was applied for a further duration of 5 sec,what would be the final velocity of the object​

Answer 1

Answer:

Explanation:

    a = Δv/t = 4/2 = 2 m/s²         and F = ma = 5 x 2 = 10 N

     we know here F is const , means a is also const

            (v - 7)/5 = 2 then , v= 17 m/s

Answer 2

Answer:

f = 5N, V = 10m/sec

Explanation:

F = m  . a

a =   (final velocity - initial velocity)/time

a =  7 - 3)/2 =  2m/sec2

F =  5*2  =    10N

now , a = 2m/sec2

time for force applied =  5sec

dv  =  a*t

dv    =  2(m/sec2)*5sec

   Final velocity  =  dv+initial velocity =  10+7 = 17m/sec.