Govt. Exams
Entrance Exams
Range = u²sin(2θ)/g = (40)²sin(90°)/10 = 1600×1/10 = 160 m
Since 5² + 12² = 25 + 144 = 169 = 13², these form a right triangle. The angle between 5 N and 12 N is 90°
Component down = mg sin30° = 100×0.5 = 50 N. Max static friction = μN = 0.5×100cos30° = 43.3 N. Since 50 > 43.3, it should slide. Let me recalculate: friction = 0.5×86.6 = 43.3 N. Actually 50 N > 43.3 N, so it WILL slide.
a = dv/dt = 6t - 6. Setting a = 0: 6t - 6 = 0, t = 1 s
Net force = √(3² + 4²) = 5 N. Using F = ma, a = 5/2 = 2.5 m/s²
Max height = u²sin²θ/(2g) = (20)²(sin45°)²/(2×10) = 400×0.5/20 = 10 m
Using a = (m₂ - m₁)g/(m₁ + m₂) = (5-3)×10/(5+3) = 20/8 = 2.5 m/s²
For friction-free circular motion: tan θ = v²/(rg) = (10)²/(50 × 10) = 100/500 = 0.2. Therefore θ = tan⁻¹(0.2)
At height h: KE = ½m(v₀² - 2gh) and PE = mgh. When KE = PE: ½m(v₀² - 2gh) = mgh. ½v₀² - gh = gh. ½v₀² = 2gh. h = v₀²/4g
Using F = ma, first find acceleration: a = (v - u)/t = (0 - 30)/0.5 = -60 m/s². Force F = ma = 1500 × 60 = 90000 N (magnitude)
