Question

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

Answer 1

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

Answer 2

Answer:

92,259.73 Joules

Explanation:

α=12.5°

m=1000 kg

d= 45m

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

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

$F= 2,123.27 N$

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

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

$W= 92,259.73 Joules$