Question

a body of mass 3kg is moving with a velocity of 2m/s.a constant horizontal force acts on the body for 3sec.so that it's velocity becomes 8 m/s.find the force​

Answer 1

Given :-

• Mass (m) of body = 3 kg
• Initial (u) velocity = 2 m/s
• Final (v) velocity = 8 m/s  
• Time (t) of force applied = 3 s

To find :-

• The force applied on the object

Answer :-

Using the First equation of motion, we can calculate acceleration (a) as,

$\boxed{v = u + at}$

$\rightarrow a = \frac{v-u}{t}$

$\rightarrow a = \frac{8-2}{3}$

$\rightarrow a = \frac{\cancel{6}}{\cancel{3}}$

$\Rightarrow a = 2 m/s^{2}$

Now, using Newton's second law of motion, we can calculate force (F) as,

$\boxed{F = m\times{a}}$

$\rightarrow F = 3\times{2}$

$\Rightarrow F = 6 N$

$\therefore$ The Force applied on the object was 6 N

Answer 2

Answer:

Step-by-step explanation:

Answer is 6N