Physics for Medical Students Exam

Name Student Number: Instructor: Section: Constants: g 9.8 m/s², 1 atm 1.013×10 Pa, Pu 1.0 10 kg/m 1) A 5.0-kg object is pulled along a horizontal surface at a constant speed by a 15-N force acting 20 above the horizontal. How much work is done by this force as the object moves 6.0 m² a) 851 b) 82 J c) 741 d) 783 c) 43J 2) When a hall rises vertically to a height h and returns to its original point of projection, the work done by the gravitational force is a) +mgh b) mgh c) 0 d) -2mgh c) +2mgh 3) A skier weighing 0.70 kN goes over a frictionless circular hill as shown. If the skier's speed at point A is 9.2 m/s, what is his speed at the top of the hill (point B)? B A 10 m 45 a) 3.1 m/s b) 5.2 m/s c) 6.5 m/s d) 4.1 m/s c) 6.2 m/s 4) An all-terrain vehicle of 2000 kg mass moves up a 15.0° slope a distance of 48 m at a constant velocity in 8 sec. The rate of change of gravitational potential energy with time is a) 30.4 kW b) 5.25 kW c) 24.8 kW d) 118 kW e) 439 kW 5) Find the pressure in atmospheres in the water at the base of a dam if the water in the reservoir is 200 meters deep. a) 194 b) 24.7 c) 29.4 d) 20.4 c) 75 6) A balloon is filled with 200 m² of helium. How large a mass can the balloon lift while moving upward at constant speed? The density of helium 0.179 kg/m³ and of air is 1.29 kg/m³. Consider the mass of the skin of the balloon to be negligible, (ignore the buoyant force on the load) a) 115 kg b) 315 kg c) 222 kg d) 415 kg e) 37 kg 7) The figure shows a uniform, horizontal beam (length 10 m, mass 25 kg) that is pivoted at the wall, with its far end supported by a cable that makes an angle of 51" with the horizontal. If a person (maas 60 kg) stands 3.0 m from the pivot, what is the tension in the cable? a) 0.83 kN b) 0.30 kN c) 0.42 KN d) 3.0 kN e) 0.38 kN 8) How much force (F) must the biceps muscle exert when a 5.0-kg mass is held in the hand with the arm horizontal as in the figure. Assume that the mass of forearm and hand together is 2.0 kg and their CG is as shown. a) 800 N d) 200 N b) 400 N e) 50 N €) 100 N 5.0 CHA (5.0 kg (8) 15cm 35cm 9) What fraction of an iceberg is submerged if p koe =917 kg/m^ 2 and mathcal D tax =1.03*10^ 1 kg/m^ 1/2 a) 77% b) 89% c) 91% d) 93% e) 95% 10) Water is flowing at 4.0 m/s in a circular pipe. If the diameter of the pipe decreases to 1/2 its former value, what is the velocity of the water downstream? a) 1.0 m/s b) 2.0 m/s c) 4.0 m/s d) 8.0 m/s c) 16 m/s 11) Water pressurized to 3.5 * 10 ^ 5 | is flowing at 5.0 m/s in a horizontal pipe which contracts to 1/3 its former area. What are the pressure and velocity of the water after the contraction? a) 4.5 * 10 ^ 4 * Pa, 1.5m / s d) 2.5 * 10 ^ 3 b) 3 * 10 ^ 5 * 1 , 10 m/s 0) 5.5 * 10 ^ 5 . 1.5 m/s c) 3 = 10 ^ 2 Pa. 15 m/s 12) The diagrams below show forces applied to a wheel that weighs 20 N. The symbol W stands for the weight. In which diagram(s) is (are) the wheel in static equilibrium? (the wheel is NOT pivoted) 20 N 10 N 10 N 10 N 20 N 10 N 10 N 10 N 10 N 10 N B b) B C D c) C d) D e) Band C List your final answers in this table. Only the answer in this table will be graded.. Question Q1: Q2: Q3: Q4: Q5: Q6: Q7: Q8: Q9: Q10: Q11: Q12: Final Answer **Physics for Medical and Dentistry Students** **Second Exam 22/4/2015** **Solutions** **Q1)** W₂ = (F cosθ)(6) = 85 J **Q2)** Vertical displacement = 0 => Wg = 0 **Q3)** No friction => DK + DU = 0 ½m(U₂² - U₁²) + mgΔH = 0 ΔH = 10 - 10cos45 = 2.93 m ½U₂² = ½(9.2)² - gΔH U₂ = [(9.2)² - 2gΔH]½ ≈ 5.2 m/s **Q4)** P = Fv Constant velocity => F = mgsin15° P = (mgsin15°)(v) = (2000x9.8xsin15°)(4.8) = 30.4 kW Note: As speed is constant, all the work is converted into potential energy. Alternatively, ΔU = mgh = 2000x9.8x(48sin15°) P = ΔU/Δt = ΔU/8 = 30.4 kW 5) P = Pa + ρgh = 1 atm + 1000 * 9.8 * 200 / 1.013 * 10^5 atm = 1 atm + 19.35 = 20.4 atm [Image of a diagram showing a column of water with height 200m] 6) constant speed => Dynamic equilibrium ΣF = 0 ↑ FB - mg - m₂g = 0 ↑ ρₐVg - ρHeVg = m₂g (ρₐ - ρHe)Vg = m₂ m₂ = 222 kg 7) 60(T sin51°) - 60g(5) - 25g(5) = 0 T = 180g + 125g / 10sin51° ≈ 0.38 kN