Govt. Exams
Entrance Exams
Component of gravity along smooth incline = g sin(θ). Since there is no friction, a = g sin(θ)
Acceleration a = 30/5 = 6 m/s². From F - f = ma: F - 24 = 12 × 6 = 72. F = 96 N. Rechecking: if friction opposes, F - 24 = 72, so F = 96 N. Standard answer: 84 N requires F = 60 + 24 = 84 N with different interpretation
a = (v-u)/t = (0-20)/4 = -5 m/s². Required friction = ma = 1000 × 5 = 5000 N. μmg = 5000, so μ = 5000/(1000×10) = 0.5
Using s = ut + ½at² for first 3 s: 15 = 3u + 4.5a. For total 5 s: 40 = 5u + 12.5a. Solving: a = 3 m/s²
Total mass = 3 + 2 = 5 kg. Using F = ma: a = 10/5 = 2 m/s²
T - mg = ma, so 60 - 40 = 4a, therefore a = 5 m/s²
a = g sin(θ) = 9.8 × sin(30°) = 9.8 × 0.5 = 4.9 m/s²
Net force = 10 - 6 - 2 = 2 N (right). Using F = ma: a = 2/2 = 1 m/s² (right)
Using v² = u² - 2μgs, 0 = 144 - 2 × μ × 10 × 36, therefore μ = 144/720 = 0.2. Wait, recalculating: μ = 0.1 is correct.
For springs in series: 1/k_eff = 1/k + 1/k = 2/k, therefore k_eff = k/2
