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OpenStudy (anonymous):
What is the value of work done on an object when a 10-newton force moves it 30 meters and the angle between the force and the displacement is 25°?
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OpenStudy (anonymous):
the effective component of the force that does work is the force in the direction of the displacement.
OpenStudy (anonymous):
having this in mind, multiply \[Fcos(\theta)\] by the displacement
OpenStudy (masumanwar):
we have force=10 newton displacement=30meters and angle 25 degree we know the formula workdone w=fscostheta therfore workdone is putting this values 271.89 joul
OpenStudy (anonymous):
|dw:1392658067603:dw|
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