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Grade 12th passMechanics

While a 1700-kg automobile is moving at a constant speed of 15 m/s, the motor supplies 16 kW of power to overcome friction, wind resistance, and so on. {a) What is the effective retarding force associated with all the frictional forces combined? (b) What power must the motor supply if the car is to move up an 8 .0 % grade (8 .0 m vertically for each 10 0 m horizontally) at 15 m/s? (c) At what downgrade, expressed in percentage terms, would the car coast at 15 m/s?

Profile image of Ashhfaq
6 Years agoGrade 12th pass
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1 Answer

Profile image of Harsh
6 Years ago
We know that power is equal to force*velocity
Hence,
16000=f*15
F=16000/15
=1066.6  N
For vertical grade.
V²=2as
225=2a*10
a=2.25/2
a=1.125 m/s²
In moving vertically upwards it there will be weight
force acting downwards and the vehicle will try moving upwards
Balancing the forces
mg+ma=Fnet
m(g)+a =Fnet
Fnet=18912 N
Power=f*v
Power=18912*15
Power =283.68 KW
Now for downgrade follow the same step in reverse order to find new displacement