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Physics 27 Online
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°?

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):

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