Question

A block of mass 50 kg is pushed on a frictionless inclined plane by a horizontal force F. The incline makes an angle of \(60^\circ\) with the horizontal. If the block moves with constant speed, then F is: (Take \(g=10\, m\,s^{-2}\))

• A. \(\frac{1000}{\sqrt{3}}\,N\)
• B. \(866\,N\)
• C. \(654\,N\)
• D. \(500\sqrt{2}\,N\)

Hint:

For incline problems with horizontal force, always resolve forces along the plane using the angle between force direction and incline.

Answer 1

The Correct Option is B

Step 1: Force resolution.
On a frictionless 60° incline, weight component down the plane = mg sin60° = 50×10×(√3/2) = 250√3 N.

Step 2: Horizontal force component.
The applied horizontal force F has an uphill component F cos60° = F/2.

Step 3: Constant speed condition.
Zero net force along the incline: F/2 = 250√3.

Step 4: Solving for F.
F = 500√3 ≈ 866 N.

Step 5: Conclusion.
The necessary horizontal force is approximately 866 N.