Calculating the Force from Power Loss at 75.0 Km/hr

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At a speed of 75.0 Km/hr, 63.0% of a car's engine power is used to overcome road and wind resistance. The engine output is 69.00 HP, equating to 51,474 Watts. The user attempted to calculate the force but made errors in their calculations and did not provide complete details of their method. It's suggested to check the units used in the calculations for consistency and clarity. Providing more detailed calculations would facilitate better assistance in determining the correct force in Newtons.
jacksondwrd
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You are testing a new car in a wind-tunnel for road and wind resistance. At a speed of 75.0 Km/hr you have found that 63.0 % of the total output power of an automobile engine is used in overcoming the resistance of the road and wind against the movement of the car. If the output power of a particular car engine is 69.00 HP, what is this resistance (ie. what is the force)? Give your answer in Newtons. (Take one HP as 746 Watts.


so i was like calcualting it like this

69HP=51474
and 60% of 51474 is 32428.62

then i did the formula of the power

power=forcexdistance/time

and got that answer but it is wrong

can anyone explain to me what i did wrong here
 
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Well, you stopped showing your calculations after figuring the power used to overcome wind resistance. I don't know what you did to calculate the force, but did you check the units of your calculations all the way thru?
 
As SteamKing mentione, you didn't mention what units you used for the terms in power = force x speed. It would help us to help you if you showed more of your work.
 

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