Govt. Exams
Entrance Exams
At the highest point: T + mg = mv²/r. When T = 500 N: 500 + 20 × 10 = 20 × v²/2. 500 + 200 = 10v². 700 = 10v². v² = 70. v ≈ 8.37 m/s. Hmm, closest to 10? Let me recalculate: 500 = 20v²/2 - 200. 700 = 10v². v = √70 ≈ 8.37 m/s. Actually checking: if v = 5, then T = 20 × 25/2 - 200 = 250 - 200 = 50 N. Not 500. Let me use: 500 + 200 = 20v²/2, so 10v² = 700, v² = 70, v ≈ 8.37. But option A is 5. Let's verify: the answer should be closer to 7-8 m/s.
Horizontal component: Fₓ = 30 cos(20°) ≈ 28.2 N. Vertical component: Fᵧ = -30 sin(20°) ≈ -10.3 N (downward). Normal force: N = mg + 10.3 = 50 + 10.3 = 60.3 N. Friction: f = 0.3 × 60.3 ≈ 18.1 N. Net force: F_net = 28.2 - 18.1 ≈ 10.1 N. Acceleration: a ≈ 10.1/5 ≈ 2 m/s². Closest to 1.8 m/s².
Using work-energy theorem: W = ½m(v_f² - v_i²) = ½ × 2000 × (5² - 20²) = 1000 × (25 - 400) = 1000 × (-375) = -375000 J. Wait, that matches option A. Let me recalculate: W = ½ × 2000 × (25 - 400) = 1000 × (-375) = -375000 J.
Maximum static friction on 6 kg block = μ × m × g = 0.3 × 6 × 10 = 18 N. If blocks move together: a = 32/(6+10) = 2 m/s². Force needed for 6 kg block = 6 × 2 = 12 N < 18 N. So blocks move together with a = 2 m/s². But checking: friction available = 18 N can support up to a = 18/6 = 3 m/s². System acceleration = 32/16 = 2 m/s². The 6 kg block needs 12 N which is available, so it moves at 2 m/s². Rechecking: If 10 kg block alone: a₁₀ = 32/10 = 3.2 m/s². Max acceleration for 6 kg = 18/6 = 3 m/s². They slip. 6 kg block accelerates at 18/6 = 3 m/s².
