Question

an object of mass 7kg moving with a velocity 14 m /sec a force is applied so that in 28 sec it attain velocity of 35m /sec what is the force applied on the objecti​

Answer 1

Answer :-

Given :-

• Mass = 7 kg
• Initial velocity = 14 m/s
• Time = 28 sec
• Final velocity = 35 m/s

To Find :-

• Force applied

Solution :-

To find force, we first need to calculate acceleration -

Calculating Acceleration :-

• u = 14 m/s
• t = 28 s
• v = 35 m/s

Substituting the value in 1st equation of motion :-

➩ $\sf v = u + at$

➩ $\sf 35 = 14 + 28a$

➩ $\sf 28a = 21$

➩ $\sf a = \frac{21}{28}$

➩ $\sf a = 0.75 m/s^2$

Calculating Force :-

• a = 0.75 m/s²
• m = 7 kg

➩ $\sf Force = Mass \times Acceleration$

➩ $\sf = 0.75 \times 7$

➩ $\sf = 5.25$

Force = 5.25 N

Answer 2

Given :

• Mass of the object = 7 kg
• Initial velocity = 14 m/s
• Final velocity = 35m/s
• Time taken = 28 s

To find :

• Force applied

Solution :

Let us use the first equation of motion in order to find acceleration.

$\implies\sf{v = u + at }$

here,

• v = final velocity
• u = initial velocity
• a = acceleration
• t = time

⇒ 35 = 14 + a x 28

⇒ a = 35 - 14 / 18

⇒ a = 21/28

⇒ a = 0.75 m/s²

Now,

By using Newton's second law,

$\implies\sf{F = ma}$

here,

• F = force
• m = mass
• a = acceleration

⇒ F = 7 x 0.75

⇒ F = 5.25 N

Hence,

The force applied to the object is 5.25N.