Question

A constant force acts on an object of mass 5kg for 2s. It increases objects velocity from 3m/s to 7m/s. calculate applied force

Answer 1

f=ma
since,
a=v-ut
given..
mass=5kg
time=2 secs
final velocity=7m/s
initial velocity=3m/s
by the sum----
f=m(v-u/t)
 =5(7-3/2)
 =5*2
 =10 N
HOPE it helps u.............

Answer 2

$\huge\mathcal\purple{Heya..Mate!!}$

$\huge\bold\pink{ANSWER:-}$

We have been given that u= 3 m/s and v= 7 m/s , t= 2s and m= 5kg.

We have the equation,

F= m(v-u)/t

Substitution of values in this relation gives,

F= 5kg (7m/s - 3m/s) / 2s = 10 N

Therefore, the applied force = 10 N

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Hope it helps...:-)

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