Q. Force of 50 N acting on a body at an angle θ with horizontal. If 150 J work is done by displacing it 3 m, then θ is
1) 60°
2) 30°
3) 0°
4) 45°
Answer:
3) 0°
Solution:
The horizontal component of force = Fcos θ
Work done = (Fcos θ) x 6
150 = 50 x cos θ x 3
cos θ = 1
θ = 0°