Question

A car of mass 500kg is moving at a speed of 36km/h is stopped by applying brakes in 10sec. calculate the force applied by the brakes. ​

Answer 1

Given :

▪ Mass of the car = 500kg

▪ Initial velocity = 36kmph

▪ Final velocity = zero

(i.e., it is brought to rest)

▪ Time interval = 10s

To Find :

▪ Force applied by the brakes.

Concept :

↗ Applied Brakes will produce uniform retardation in the car and finally car will come to rest.

↗ We can calculate force applied by brakes by Newton's second law of motion.

↗ As per this law, Force is defined as the product of mass and acceleration.

Conversion :

⭐ 1kmph = 5/18mps

⭐ 36kmph = 36×5/18 = 10mps

Calculation :

✴ Acceleration of car :

$\dashrightarrow\sf\:v=u+at\\ \\ \dashrightarrow\sf\:0=10+a(10)\\ \\ \dashrightarrow\sf\:10a=-10\\ \\ \dashrightarrow\bf\:\red{a=-1\:ms^{-2}}$

⚠ -ve sign shows retardation.

✴ Force applied by brakes :

$\twoheadrightarrow\sf\:F=ma\\ \\ \twoheadrightarrow\sf\:F=500\times 1\\ \\ \twoheadrightarrow\underline{\boxed{\bf{\blue{F=500N}}}}\:\orange{\bigstar}$

☣ Force acts in opposite to the motion.

Answer 2

Given

• Mass of car = 500 kg
• Initial speed (u) = 36 km/h
• Time to stop (t) = 10 s

To find

• Force applied by the brakes.

Solution

From the data ;

Mass (M) = 500 kg

Initial speed (u) = 36 km/h

= 36 × (5/18) m/s

= 10 m/s

Final speed (v) = 0 m/s

Time (t) = 10 s

Now using 1st equation of motion :

→ v = u + at

Substituting values,

→ 0 = 10 + a × 10

→ 0 - 10 = 10a

→ 10a = -10

→ a = -10/10

→ a = -1 m/s²

Hence, acceleration of car = -1 m/s²

Now using Newton's 2nd law :

→ Force = Mass × Acceleration

Substituting values,

→ Force = 500 × (-1)

→ Force = -500 N

[Negative sign shows that force is acting against the motion of car]

Hence, 500 N force is applied by brakes against the motion of car.