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Question:

What is the minimum work needed to push a 950 kg car 930 m up along a 8.0 degree incline?

Assume the effective coefficient of friction retarding the car is 0.22.(i solved the problem multiple times and got 1.5*10^6 J, but that is not the correct answer. I want to know what I did wrong) Thank you.

Answer:

does the check engine light blink on and off, or does it just just say on? maybe it your cat is no good (clogged) and the o2 is reading it wrong. the muffler shop near you can check it out. if the light blinks, that is a sign of a cat going bad.
does the check engine light blink on and off, or does it just just say on? maybe it your cat is no good (clogged) and the o2 is reading it wrong. the muffler shop near you can check it out. if the light blinks, that is a sign of a cat going bad.
First calculate the height that the car reaches above the starting point: height 930*sin 8° height 129m Work done in lifting car to this height m*g*h Work 950*9.81*129 Work 1,202,215.5J Now you have to account for the friction: Coefficient of friction 0.22. Frictional force retarding motion 0.22m*g*cos 8° friction force 0.22*950*9.81*0.990 Frictional force 2,029.8N Work done against friction Force * distance Work done against friction 2029.8*930 1,887,702J Total work done 1,202,215.5 + 1,887,702 3,089,918J Work required 3.1*10^6J
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First calculate the height that the car reaches above the starting point: height 930*sin 8° height 129m Work done in lifting car to this height m*g*h Work 950*9.81*129 Work 1,202,215.5J Now you have to account for the friction: Coefficient of friction 0.22. Frictional force retarding motion 0.22m*g*cos 8° friction force 0.22*950*9.81*0.990 Frictional force 2,029.8N Work done against friction Force * distance Work done against friction 2029.8*930 1,887,702J Total work done 1,202,215.5 + 1,887,702 3,089,918J Work required 3.1*10^6J

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