The driver of a 1350 kg car, coasting down a hill, sees a red signal at the bottom, for which he must stop. His speed at the time the brakes are applied is 28 m / sec. and he is 30 m vertically above the bottom of the hill. How much energy as heat must be dissipated by the brakes if wind and other frictional effects are neglected ? Take g as the standard acceleration of gravity equal to 908.66 m / s2
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Answers
Answer: It's a little strange to answer a question .about energy dissipated in braking “neglecting frictional effects” since braking is a frictional effect. The answer should be zero if frictional effects are neglected.
I assume what you're really asking is how much energy did the vehicle have initially?
The vehicle has two energy componentskinetic energy since it's moving and potential energy because it's above the bottom of the hill. To bring the vehicle to a stop at the bottom all of that energy must be dissipated.
The kinetic energy is 0.5*m*v^2, 270,000 Joules. The potential energy is m”g*h which is 794,610 J. So the total energy is 1064610 J.
If I multiply by 4.182 I get the energy in calories, 4452199 c. That much energy would bring 55.65 kg of water from.20 C to the boiling point, 100C. That's a cubical volume of water 38.2 cm on a side. That's figuring that a gram of water is 1 cubic centimeter..
Explanation: