Question

A constant force applied on an object of mass 2 kg for 10 s, increases its velocity from 5 m/s to 10 m/s. Find the magnitude of applied force. If the same force is applied for 15 s, what will be the final velocity of the object
PLEASE answer ​

Answer 1

Answer:

M=2kg=

T=10sec

u=5m/sec

v=10m/sec

v=u+at

a=(v-u)/t=(10-5)/10=1/2m/sec²

F=Ma=2×1/2=1/kg/sec²=1

If force is applied for 15sec then velocity is

F=mv/t

v=(1/15)/2=7.5m/sec

Have a great day dear ✌️........

Answer 2

m=2 kg

t= 10 s

u= 5 m/s

v= 10 m/s

F= ?

we have

a= (v-u)/t

= (10-5)/10

=0.5 m/s²

now

F= ma

= 2* 0.5

= 1 N

Again

same force is applied in same mass body

so a = 0.5 m/s²

t= 15 sec

u= 5m/s

v= ?

we have

v= u + at

= 5 + 0.5*15

=. 12.5 m/s²