Your 1500-kg car, moving at 7.0 m/s, approaches the bottom of a hill that is 20
ID: 1363982 • Letter: Y
Question
Your 1500-kg car, moving at 7.0 m/s, approaches the bottom of a hill that is 20 m high (Figure 1) . To save gas, you use on average only 3.1 kW of engine power, realizing that half of the energy delivered by the engine and half of the initial kinetic energy will be dissipated. If you calculated correctly, your car will just barely make it over the hill.
What time interval is required for your car to travel from the bottom to the top of the hill?
It is not: 47s, 78s, 83s, 95s, or anything that would round to that.
Explanation / Answer
the time is
t = U2- U2/ P
= mgh + KE1/2 - KE1/P
= mgh - KE1/P
=1500( 9.8) (20 ) - 1/2 * 1500 ( 7)^2/ 3100 W
=82.98 s
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