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)

• A. 2 m s–2
• B. Zero
• C. 0.1 m s–2
• D. 1 m s–2

Answer 1

Answer: (D)

1 m s–2

Answer 2

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