Question:

A box of mass 5 kg is pulled by a cord, up along a frictionless plane inclined at 30° with the horizontal. The tension in the cord is 30 N. The acceleration of the box is (Take g = 10 m s–2)

Updated On: Dec 9, 2024
  • 2 m s–2
  • Zero
  • 0.1 m s–2
  • 1 m s–2
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The Correct Option is D

Solution and Explanation

Let's analyze the forces acting on the box:

  • Tension (T) = 30 N (up the incline)
  • Weight (mg) = 5 kg × 10 m/s2 = 50 N (vertically downwards)

The component of weight parallel to the incline is mg sin30° = 50 × $\frac{1}{2}$ = 25 N (down the incline).

Applying Newton's second law along the incline (upward direction as positive):

T − mg sin 30° = ma

30 N − 25 N = 5 kg × a

a = $\frac{5}{5}$ = 1 m s-2

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