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Hunter-Best [27]

3 years ago

10

Assuming no friction, what is the minimum work needed to push a 1000 kg car 45.0 m up a 12.5 incline?

2 answers:

Dvinal [7]3 years ago

6 0

Work=Fdcos(theta) so (10000)(45.0)cos(12.5 degrees) = 440000 J

bulgar [2K]3 years ago

4 0

Answer:

92,259.73 Joules

Explanation:

α=12.5°

m=1000 kg

d= 45m

$F=m*g*Sin(\alpha)\\$ °) $F= 1000kg*9.81 \frac{m}{s} *Sin (12.5 )\\$

$F= 1000 kg * 9.81 \frac{m}{s} * 0.2164\\$

$F= 2,123.27 N\\$

$W=F*d*Cos (\alpha)\\$

$W= 2,123.27 N * 45m * Cos (12,5)\\$

$W= 92,259.73 Joules$

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