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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman
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Automotive Power II. (a) If 8.00 hp are required to drive

Chapter 3, Problem 101P

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QUESTION:

Automotive Power II. (a) If 8.00 hp are required to drive a 1800-kg automobile at 60.0 km/h on a level road, what is the total retarding force due to friction, air resistance, and so on? (b) What power is necessary to drive the car at 60.0 km/h up a 10.0% grade (a hill rising 10.0 m vertically in 100.0 m horizontally)? (c) What power is necessary to drive the car at 60.0 km/h ?down a 1.00% grade? (d) Down what percent grade would the car coast at 60.0 km/h?

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QUESTION:

Automotive Power II. (a) If 8.00 hp are required to drive a 1800-kg automobile at 60.0 km/h on a level road, what is the total retarding force due to friction, air resistance, and so on? (b) What power is necessary to drive the car at 60.0 km/h up a 10.0% grade (a hill rising 10.0 m vertically in 100.0 m horizontally)? (c) What power is necessary to drive the car at 60.0 km/h ?down a 1.00% grade? (d) Down what percent grade would the car coast at 60.0 km/h?

ANSWER:

Solution 101 P Step 1: Given data Power P = 8.00 hp = 5968 Watts Mass m = 1800 kg Speed v = 60 km / h = 16.66 m/s We need to find the total retarding force due to friction , air and other factors We know P = F × v This can be rearranged to find force F = 5968 Watts / 16.66 m/s F = 358 N Hence we get force as 358 N

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