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Community Contributions - Articles by goIITians
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| small note in work energy |
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Work-Energy Principle
The change in the kinetic energy of an object is equal to the net work done on the object. This fact is referred to as the Work-Energy Principle and is often a very useful tool in mechanics problem solving. It is derivable from conservation of energy and the application of the relationships for work and energy, so it is not independent of the conservation laws. It is in fact a specific application of conservation of energy. However, there are so many mechanical problems which are solved efficiently by applying this principle that it merits separate attention as a working principle. For a straight-line collision, the net work done is equal to the average force of impact times the distance traveled during the impact. Average impact force x distance traveled = change in kinetic energy If a moving object is stopped by a collision, extending the stopping distance will reduce the average impact force. Example of Force on Car
(This initial example is cast in U.S. common units because most U.S. readers can make comparisons to known forces more easily in those terms. The calculation provides the results in SI units as well.)
Vary the parameters of the collision Force on Driver in Example Car Crash For the car crash scenario where a car stops in 1 foot from a speed of 30 mi/hr, what is the force on the driver? Assume a 160 lb (mass = 5 slugs) driver. If firmly held in non-stretching seatbelt harness: Stopping distance 1 ft. | | - Deceleration = 967 ft/s2 = 294 m/s2 = 30 g's
- Force = 4813 lb = 21412 N = 2.4 tons
| Non-stretching seatbelt | If not wearing seatbelt, stopping distance determined by nature of collision with windshield, steering column, etc. : stopping distance 0.2 ft. | | - Deceleration = 4836 ft/s2 = 1474 m/s2 = 150 g's
- Force = 24068 lb = 107059 N = 12 tons!!
| No seatbelt! | If seat belt harness stretches, increasing stopping distance by 50%: 1.5 ft. | | - Deceleration = 645ft/s2 = 197 m/s2 = 20 g's
- Force = 3209 lb = 14274 N = 1.6 tons
| Stretching seatbelt | These calculated numbers assume constant deceleration, and are therefore an estimate of the average force of impact.
Stopping Distance for Auto Assuming proper operation of the brakes, the minimum stopping distance for an automobile is determined by the effective coefficient of friction between the tires and the road. The friction force of the road must do enough work on the car to reduce its kinetic energy to zero (work-energy principle). If the wheels of the car continue to turn while braking, then static friction is operating, while if the wheels are locked and sliding over the road surface, the braking force is a kinetic friction force.
To reduce the kinetic energy to zero:
so the stopping distance is
Note that this implies a stopping distance independent of vehicle mass. It also implies a quadrupling of stopping distance with a doubling of vehicle speed. |
| Even though the stopping distance for a large truck should be the same as that for a car, there are real-world cases where a truck has more difficulty stopping. Runaway truck ramps are placed on long downgrades like this one on Monteagle in Tennessee. Deep, loose gravel provides a stopping force greater than that obtained from just a tire against the road. |
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