A force F = (2.75 N) i + (7.20 N) j + (6.80 N) k acts on a 2.00 kg mobile object

Question

A force F = (2.75 N) i + (7.20 N) j + (6.80 N) k acts on a 2.00 kg mobile object that moves from an initial position of di = (3.00 m) i - (2.00 m) j + (5.00 m) k to a final position of df = -(5.00 m) i + (3.85 m) j + (7.00 m) k in 4.00 s. Find the work done on the object by the force in the 4.00 interval J Find the average power due to the force during that interval. W Find the angle between vectors, dj and df. degree How do you calculate the work done by a constant force? How do you calculate the displacement of a particle given its initial and final positions? How is average power related to the work done and the elapsed time?

Explanation / Answer

a.) work = F.s

= (2.75 i + 4.20 j + 6.80 k ) . ( -8i +5.85j +2k)

= -22 + 24.57 + 13.6

= 16.17 J

b.) power = Work/time

= 4.042 W

c.) cos = AxBx + AyBy +AzBz/AB

= ( -22 +24.57 +13.6 )/(71.44*102.22)

= 2.214 *10^-3

= 89.87 degree

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