Question
Download Solution PDFA car weighs 1000 kg. It is moving with a uniform velocity of 72 km/h towards a straight road. The driver suddenly presses the brakes. The car stops in 0.2 s. The retarding force applied on the car to stop it is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFCONCEPT:
Retarding Force (Fretard)
Fretard = m x a
a = (v - u) / t
EXPLANATION:
- Given data:
- Mass of the car (m) = 1000 kg
- Initial velocity (u) = 72 km/h = 72 * (1000/3600) m/s = 20 m/s
- Final velocity (v) = 0 m/s
- Time taken to stop (t) = 0.2 s
- a = (v - u) / t
- = (0 - 20) / 0.2
- = -20 / 0.2
- = -100 m/s2
- Fretard = m * a
- = 1000 kg * (-100 m/s2)
- = -100,000 N
- The negative sign indicates that the force is in the opposite direction of the motion. The magnitude of the retarding force is:
- 100,000 N or 100 kN
Therefore, the retarding force applied on the car to stop it is 100 kN.
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