A hydraulic brake system of a car of mass 1000 kg having speed of 50 km/h, has a cylindrical piston of radius of 0.5 cm. The slave cylinder has a radius of 2.5 cm. If a constant force of 100 N is applied on the brake what distance the car will travel before coming to stop?
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Given A hydraulic brake system of a car of mass 1000 kg having speed of 50 km/h, has a cylindrical piston of radius of 0.5 cm. The slave cylinder has a radius of 2.5 cm. If a constant force of 100 N is applied on the brake what distance the car will travel before coming to stop?
- So mass m = 1000 kg
- Speed v = 50 km / hr = 50 x 5/ 18 = 250 / 18 m/s
- Now there is a circular piston of radius r1 = 0.5 cm = 0.5 x 10^-2 m
- Now the larger area slave cylinder whose radius r2 = 2.5 cm = 2.5 x 10^-2 m
- Since this is circular in shape we have area A1 = π r1^2
- = π x (0.5 x 10^-2)^2
- Also area A2 = πr2^2
- = π x (2.5 x 10^-2)^2
- Now a constant force of 100 N is applied on the brake
- So F1 = 100 N
- Now we need to find F2 that is the force applied to stop.
- So we have F2 = F1 x A 2 / A1
- = 100 x π x (2.5 x 10^-2)^2 / π x (0.5 x 10^-2)^2
- = 100 x 6.25 x 10^-4 / 0.25 x 10^-4
- = 25 x 100
- F2 = 2500 N
- So now the retarding acceleration a = F / m
- So a = 2500 / 1000
- Or a = 2.5 m /s^2
- Now the equation of motion will be
- So v^2 = u^2 – 2as (since retarding)
- So 0 = (250 / 18)^2 – 2 (2.5) s
- So s = 250 / 18 x 250 / 18 x 1/5
- Or s = 38.58 m
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