Part A How much work is done by the force F =( 1.3 i^+ 6.4 j^)N on a particle th

Question

Part A

How much work is done by the force F =( 1.3 i^+ 6.4 j^)N on a particle that moves through displacement r = 2.7 i^m

Part B

How much work is done by the force F  =( 1.3 i^+ 6.4 j^)N on a particle that moves through displacement r = 2.7 j^m?

Part C

A worker lifts a 20.0-kg bucket of concrete from the ground up to the top of a 25.0-m tall building. The bucket is initially at rest, but is traveling at 4.0 m/s when it reaches the top of the building. What is the minimum amount of work that the worker did in lifting the bucket?

A worker lifts a 20.0-kg bucket of concrete from the ground up to the top of a 25.0-m tall building. The bucket is initially at rest, but is traveling at 4.0 m/s when it reaches the top of the building. What is the minimum amount of work that the worker did in lifting the bucket?

Explanation / Answer

Part A: F = -1.3i+6.4 j, r = 2.7 i

W =F.r = [-1.3i+6.4j].(2.7i)

W = (-1.3*2.7)

W= -3.51J

Part B: F = -1.3i+6.4 j, r = 2.7 j

W =F.r = [-1.3i+6.4j].(2.7j)

W = (6.4*2.7)

W= 17.28 J

Part C: Correct option is (B)

m =20 kg, h =20 m, u =0, v =4 m/s

Work doen by non conservative force = Ef+Ei

= (Kf+Uf) -(Ki+Ui)

= (1/2)mv2+mgh

= [(1/2)(20*4*4)]+[20*9.8*25]

W= 5.07 kJ

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