Calculate the work done by the brakes of a car of mass 1000 kg when its speed is reduced from 20 m/s to 10 m/s?
Calculate the work done by the brakes of a car of mass 1000 kg when its speed is reduced from 20 m/s to 10 m/s?
Data:
Mass of a car = 1000 kg
We have to calculate the work done by the brakes when the speed of the car is reduced from 20 m/s to 10 m/s.
Let,
v₁ = 20 m/s
v₂ = 10 m/s
The kinetic energy of the car when its speed is 20 m/s is given as:
K₁ = 1/2 m v₁²
K₁ = 1/2 (1000) (20)²
K₁ = 2 x 10⁵ J
The kinetic energy of the car when its speed is reduced to 10 m/s is given as:
K₂ = 1/2 m v₂²
K₂ = 1/2 (1000) (10)²
K₂ = 5 x 10⁴ J
So, the work done by the brake will be the difference of their kinetic energies.
Therefore,
K₂ - K₁ = 5 x 10⁴ - 2 x 10⁵
K₂ - K₁ = - 150,000 J
So, the work done by the brakes to reduce the speed of the car from 20 m/s to 10 m/s is (- 150,000) J.
It's negative because the brakes oppose the motion of the car.
Data:
Mass of a car = 1000 kg
We have to calculate the work done by the brakes when the speed of the car is reduced from 20 m/s to 10 m/s.
Let,
v₁ = 20 m/s
v₂ = 10 m/s
The kinetic energy of the car when its speed is 20 m/s is given as:
K₁ = 1/2 m v₁²
K₁ = 1/2 (1000) (20)²
K₁ = 2 x 10⁵ J
The kinetic energy of the car when its speed is reduced to 10 m/s is given as:
K₂ = 1/2 m v₂²
K₂ = 1/2 (1000) (10)²
K₂ = 5 x 10⁴ J
So, the work done by the brake will be the difference of their kinetic energies.
Therefore,
K₂ - K₁ = 5 x 10⁴ - 2 x 10⁵
K₂ - K₁ = - 150,000 J
So, the work done by the brakes to reduce the speed of the car from 20 m/s to 10 m/s is (- 150,000) J.
It's negative because the brakes oppose the motion of the car.
