Question

the same amount of force is applied for 4s in an object with mass of 10 kg .calculate the force applied if the velocity of the object increases from 6m/s to 14 m/s​

Answer 1

Given:-

• Time taken ,t = 4s

• Mass,m = 10 kg

• Initial velocity ,u = 6m/s

• Final velocity ,v = 14m/s

To Find:-

• Force ,F

Solution:-

Firstly we calculate the acceleration of the object.

As we know that Acceleration is defined as the rate of change of velocity at per unit time.

• a = v-u/t

where,

v is the final velocity

a is the acceleration

u is the initial velocity

t is the time taken

Substitute the value we get

→ a = 14-6/4

→ a = 8/4

→ a = 2m/s²

Acceleration is 2m/s²

Now, Calculating the Force

Force is the product of mass and acceleration.

• F = ma

Substitute the value we get

→ F = 10 ×2

→ F = 20 N

Therefore,the force applied is 20 Newton .

Answer 2

Answer:

Given :-

• The same amount of force is applied for 4 second in an object with mass of 10 kg.
• The velocity of the object increases from 6 m/s to 14 m/s.

To Find :-

• What is the force.

Formula Used :-

❶ To find acceleration we know that,

$\boxed{\bold{\large{a\: =\: \dfrac{v - u}{t}}}}$

where,

• a = Acceleration
• v = Final velocity
• u = Initial velocity
• t = Time

❷ To find force we have to use Newton's Second law of motion,

$\boxed{\bold{\large{F\: =\: ma}}}$

where,

• F = Force
• m = Mass
• a = Acceleration

Solution :-

Given :

• Final velocity (v) = 14 m/s
• Initial velocity (u) = 6 m/s
• Mass (m) = 10 kg
• Time (t) = 4 second

❶ First we have to find the acceleration,

↦ a = $\dfrac{14 - 6}{4}$

↦ a = $\dfrac{8}{4}$

↦ a = $\sf\dfrac{\cancel{8}}{\cancel{4}}$

➠ a = 2 m/s²

Hence, acceleration is 2 m/s²

❷ Now, we have to find the force,

We get,

• Mass = 10 kg
• Acceleration = 2 m/s²

⇒ F = 10 × 2

➽ F = 20 N

$\therefore$ The force applied is 20 N .