Question

A constant force acts on an object of mass 5kg for a duration of 2s.it increase the velocity of object from 4ms-1 to 8ms-1 find the magnitude of the applied force

Answer 1

m=5kg
time taken is 2s
u=4ms-1
v=8ms-1
F=ma
F=5*(8-4/2)
F=5*(4/2)
F=5*2
F=10
hope it will help u

Answer 2

v=8m/s
u=4m/s
t=2s
m=5kg

a=v-u/t
a=8-4/2
a=2m/s^2

we know that
F=ma

F=5*2
F=10N