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What is the minimum work needed to push a 1025 kg car 290 m up a 17.5° incline with zero friction?
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Answered by
Damon
weight force down slope = m g sin 17.5
= 1025 (9.81) sin 17.5
= 3023 N
distance = 290 in direction of needed force
so
Work = 3024 N * 290 m = 876,865 Joules
By the way
if you do it in 10 minutes which is 600 seconds
that is 876,865/600 = 1,461 watts
You need about 2 horsepower :)
= 1025 (9.81) sin 17.5
= 3023 N
distance = 290 in direction of needed force
so
Work = 3024 N * 290 m = 876,865 Joules
By the way
if you do it in 10 minutes which is 600 seconds
that is 876,865/600 = 1,461 watts
You need about 2 horsepower :)
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