Question

The work done in pulling up a block of wood weighing 2kN for length of 10m on smooth incline plane at angle 15 degree with horizontal

Answer 1

Work done is equal to Force acting on a load multiplied by the displacement of the load in the direction of Force. Work is done against gravity (vertical direction)

The component of displacement in vertical direction: h cos(90- Ф)
       = 10 meters * Cos(90- 15) deg  = 2.588 meters
Work done = Force * displacement = 2 kN * 2.588 meters = 5.176 kN-m or Joules

Answer 2

Answer:

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