Question

car of mass 05 quintal moving with a speed of 54 km/hr is

stopped by applying break in 10 seconds then find force applied

by brakes.​

Answer 1

Given :

• Mass (m) = 0.5 quintal
• Initial velocity (u) = 54 km/h
• Time (t) = 10 seconds
• Final velocity (v) = 0 m/s

To find :

• Force

According to the question,

Note :

At first we will change 0.5 quintal into kg.So, it will be 50 kg.

Then,

We will change 54 km/h into m/s.So,it will be 15 m/s.

By using Newtons first equation of motion we will find acceleration then by using the formula of F = ma we will find force,

⇒ v = u + at

Where,

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

⇒ Substituting the values,

⇒ 0 = 15 + a × 10

⇒ 0 - 15 = 10a

⇒ - 15 = 10a

⇒ - 15 ÷ 10 = a

⇒ - 1.5 = a

So,the acceleration is - 1.5 m/s². Negative signs means retardation.

Now,

• Acceleration (a) = 1.5 m/s²

By using F = ma we will solve,

⇒ Force = Mass × Acceleration

Or,

⇒ F = ma

⇒ Substituting the values,

⇒ F = 50 × (-1.5)

⇒ F = - 75 N

So,the force is - 75 Newtons.

Here, negative signs means that the force is acting on opposite direction.

__________________________

Answer 2

Given -

• Mass of car = 0.5 quintal
• Initial velocity (u) = 54 km/hr
• Final velocity (v) = 0 m/s
• Time (t) = 10 seconds

To find -

• Force applied by brakes.

Solution -

• Firstly, convert the mass of the car from 0.5 quintal to kg.
• Then convert the Initial velocity from km/hr to m/s.

Mass -

As we know,

↬ 1 quintal = 100 kg

So, multiply 0.5 quintal by 100.

↬ 0.5 × 100

↬ 50 kg

• Mass = 50 kg

Initial velocity -

As we know,

↬ 1 km = 1000 m

↬ 1 hr = 3600 s

Divide the value by 3600 and multiply by 1000 to convert it into m/s.

= $\sf{\dfrac{54 \times 1000}{3600}}$

= 15 m/s

• Initial velocity = 15 m/s.

• v = u + at ( To find acceleration )

Substitute the given values.

↬ 0 = 15 + a × 10

↬ -15 = a × 10

↬ $\sf{\dfrac{-15}{10}}$ = a

↬ a = - 1.5 $\sf{m/s^2}$

Acceleration = - 1.5 $\sf{m/s^2}$

• Retardation = $\sf{m/s^2}$

• F = ma

where,

F = force

m = mass

a = acceleration

Substitute the given values.

↬ F = 50 × (-1.5)

↬ -75 N

$\LARGE{\overbrace{\underbrace{\sf{\red{Force = - 75~N}}}}}$