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COMEDK2015PhysicsRotational Motion

The diagram shows a barrel of weight 1.0 10³ ~N on a frictionless slope inclined at 30^ to the horizontal. The force is parallel to the slope. What is the work done in moving the barrel a distance of 5.0 ~m up the slope?

Options

  1. A2.5 10³ ~J
  2. B4.3 10³ ~J
  3. C5.0 10³ ~J
  4. D1.0 10³ ~J

Correct answer

A. 2.5 10³ ~J

Step-by-step solution

The given situation is shown below Work done in moving the barrel on the frictionless slope is equal to change in potential energy. i.e. W=m g (h₁-h₂ ) Here, m g=1 10³ ~N From figure, 30^ = h₁-h₂ 5 h₁-h₂=5 30^ = 5 2 =25 ~m Putting these values in Eq. (i), we get W=1 10³(2.5)=2.5 10³ ~J

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