Question

A car moves at a constant velocity of 30 m/s and has 3.6 x 10 5 J of kinetic energy. The driver applies the brakes and the car stops in 95 m.
a. Calculate the force needed to stop the vehicle

Answer 1

Answer:

3789 N

Explanation:

Kinetic energy = ½ mu²

=> 3.6 x 10^(5) = ½ m(30)²

=> 3.6 x 10^(5) x 2/30² = m

=> 800 kg = m

Using equations of motion,

v² = u² + 2aS, here v(final velocity is 0, when it comes to rest, S = 95, u = 30).

=> 0² = 30² + 2a(95)

=> - 4.7368 ≈ a , -ve represt the state that velocity is deceasing.

So force required to stop it is:

=> m x a = 800 x 4.7368 N

= 3789 N

Answer 2

$\huge {\underline{\green{Given:-}}}$

• Velocity ,v (u) = 30m/s

• Kinetic energy ,KE = 3.6×10⁵ J

• Distance ,s = 95 m

• Final velocity ,v = 0m/s

$\huge {\underline{\pink{To Find:-}}}$

• Force needed to stop the vehicles ,F

$\huge {\underline{\red{Solution:-}}}$

As we know that Kinetic Energy is the energy possed due to motion.

• KE = 1/2mv²

Substitute the value we get

→ 3.6×10⁵ = 1/2×m × 30²

→ 360000 = 1/2 × m × 900

→ 360000 = 450 × m

→ m = 36000/450

→ m = 800 kg

. Therefore, the mass of the car is 800kg

Now Calculating the acceleration of the car

Using 3rd Equation of Motion

• v² = u² +2as

Substitute the value we get

→ 0² = 30² + 2× a × 95

→ 0 = 900 + 190 × a

→ -900 = 190×a

→ a = -900/190

→ a = -4 .73 m/s²

here, Negative sign show retardation

Therefore the acceleration of the car is 4.73m/s²

Now , We know that Force is the product of mass and acceleration

• F = ma

Substitute the value we get

→ F = 800 × 4.73

→ F = 3784 N

Therefore, the force required to stop the vehicles is 3784 Newton .