A car traveling at 26m/s runs out of gas while traveling at 8 degree slope.?

The car goes on until the force of gravity brings it to a stop.


.............._/¯¯¯¯\__.Vi26m/s
.............OO___-----
____--------|\--- Wsin8°
.......................|..\
.......................|8°.\.N

The car has a Kinetic Energy directed upwards the slope.
The component of Weight of the car parallel to the slope retards the car for a distance S until Vf becomes zero (Vf = 0)

KE(final) = KE(initial) - Work(retarding force)
½mVf ² = ½mVi ² - (Wsin8°)S
0 = ½mVi ² - mgsin8°S .........m cancels
gsin8°S = ½Vi ²
S = Vi ² /2gsin8°
S =(26)² /2 (09.8)(.139)

S = 248.1 m
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Another way is to use KE = PE

………............................….… |
……......................…… ..........|
….................S… ...................|
…..........… ............................|h
….... ......8°............................|
__________________________|

½mV ² = mgh..........m cancels
V ² = 2gh.
h = V ² / 2g
h = (26)² / 2(9.8)
h = 34.49 m

sin8° = h / S

S = (34.49)/(.139)
S = 248.1m
 
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