Static Coefficient Friction - Physics - Solved Past Exam, Exams of Physics

Download Static Coefficient Friction - Physics - Solved Past Exam and more Exams Physics in PDF only on Docsity! Solution of PC1141 05-06 Physics Society 10/7/2007 1. Fd = mgsinθ Static co-efficient friction = tan30o = 0.577 mgsinθ - fk = ma Using s=ut+1/2 ·at2 formulae, 4.0m = 1 2 at2 →a = 0.5m/s ⇒ fk = mgsinθ −ma = m(gsinθ-a) Kinetic co-efficient of friction = m(gsinθ - a)/mgsinθ = gsinθ−a gcosθ = 9.81×sin30 o−0.5 9.81×cos30o = 0.518 2. • Force to cause the oscillation = mgsinθ = ma sinθ ≈ θ for small angle → lθ = x ⇒ ma = mg(x l ) as sinθ ≈ θ a = −g l · x ω2 = g l T = 1 2 × (4π2l g )0.5 + 1 2 × (4π20.5 g )0.5 = 1.71s • Note that the angular momentum is not conserved. Anyway conservation of energy still hold. Hence the highest height achieved by the pendulum is still 0.5m. 3. • Rate of work done = Power V=dx dt = 15t2-16t-30 a = d 2x dt2 = 30t− 16 Power = Fv = 0.280kg ×a× V At t = 2.0s, Power = -24.64W ignore negative as we would like to look at the magnitude At t = 4.0s, Power = 4251.5W 1 • Average Power = 1 2 ∫ 4.0 2.0 F · dx = 1 2 ∫ 4.0 2.0 (0.28) · (30t− 16) · (15t2 − 16t− 30)dt = 1 2 · 0.28 ∫ 4.0 2.0 (450t 3 − 480t2 − 644t + 480)dt = 0.14 · [ 450 4 t4 − 720 3 t3 − 322t2 + 480t ]4 2 =1490W 4. In one second, let m = mass of air strike the person. m = 1.5 × 0.5 × 150 × 1000 3600 × 1.2kg = 37.5kg → Force = mv t = 37.5kg × 150 × 1000 3600 m/s2 = 1562.5N 5. • Given the wave equation is y = 2.0cos(0.50πx− 200πt) Therefore, amplitude = 2.0cm 0.50π = 2π λ λ = 4cm 200π = 2πf f = 100Hz T = 1 f T = 0.01s Velocity = 100 × 0.04 = 4m/s • v = √ T µ T = v2µ = (4m/s)2 × 0.005 0.01 = 8N 6. • Please refer to the text book for the show of inertia of a uniform cylinder where the symmetry axis passing through its center of mass. • G. Potential energy will convert into Kinetic energy + Elastic potential energy through out the process. ⇒ 2.0 × 9.81 × 1 × sin37o = 1 2 × 2.0 × V 2 + 1 2 × 20 × (1.0)2 + 1 2 · (1 2 × 5× 0.32)× ( V 0.3 )2 1.81J = 9 4 V 2kg V = 0.9m/s • Let h = the vertical distance where the block has slided down. → h = x sin37o where x is the distance the block has slided down along the plane. ⇒ mgxsin37o = 0.5 · 20 · x2 2 Let T = Tension in the string when the puck has the minimum velocity at lowest position T - mgsinθ = m(5gLsinθ)÷ L → T = 6mgsinθ = 6× 1.2× 9.81× sin40o = 45.4N • Let the velocity at lowest point required to complete the circular path = V0 and the velocity at the highest point = Vf . T = tension of the string Based on the conservation of energy, 0.5mV 20 = 0.5mV 2 f + mgh 0.5mV 20 = 0.5mV 2 f + mg(2Lsinθ) (1) But mV 2f L = T + mgsinθ Let’s assume T = 0N when the puck is at highest point for minimum V0. → mV 2 f L = mgsinθ V 2f = gLsinθ subinto the equation (1) ⇒ 0.5mV 20 = 0.5mgLsinθ + mg2Lsinθ V 20 = 5gLsinθ At last we have the tension T0 at lowest point as mV 20 L = T0 −mgsinθ T0 = 6mgsinθ = 6× 1.2× 9.81× sin40o = 45.4N • the puck will slide following the circular path but it will not reach the top due to frictional force. • this time the string has been replaced by a rod with negligible mass. Using conservation of energy again, but this time even when the puck reaches highest point, it can even complete the circular path even without velocity. (the rod will support puck but not the string when it is at highest point). 1 2 mV 2i = mg2Lsinθ Vi = √ 4gLsinθ = √ 4× 9.81× 0.75× sin40o = 4.35m/s 5 • I think that even the rod has no negligibled mass, it will still need the same minimum velocity to complete the circular path. Because if we manage the give the rod with the same velocity as calculated in part (d), that means the rod has sufficient kinetic energy as well to complete the circular path. 6