Presentation is loading. Please wait. # Problem 2.8 An object whose mass is 1 lb has a velocity of 100ft/s. Determine the final velocity in ft/s. if the kinetic energy of the object decreases.

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Problem 2.8 An object whose mass is 1 lb has a velocity of 100ft/s. Determine the final velocity in ft/s. if the kinetic energy of the object decreases by 100 ft. lbf. Determine the change in elevation in ft. associated with a 100 ft. lbf change in potential energy. Let g = 32.0 ft/s2. m = 1 lb. V1 = 100 ft/s. (KE)1 = ½ m V12= ½ (1)(100)(100) = 5000 lbm ft2/s2. 1 ft. lbf = (1) (ft) (32.2) lbm ft/s2 KE 2 = 5000 – 32.2 x 100 = 1780 lbm. ft2/s2 V2 2 = 2 x (1780) / (1); V2 = ft/s

Problem 2.16 Beginning from rest, an object of mass 200 kg. slides down a 10-m long ramp. The ramp is inclined at an angle of 40 deg. From the horizontal. If air resistance and friction between the object and ramp are negligible, determine the velocity of the object in m/s at the bottom of the ramp. g = 9.81 m/s2 Change in elevation = 10 sin (40) = 10(0.643) = m. DKE = - mg Dz; KE 2 = ½ m (V2)2 + 0 = -m (9.81) (-6.43) (V2)2 = (2) (9.81) (6.43) = m2/s2 V2 = m/s

Problem 2.19 An object of mass 80 lb. initially at rest experiences a constant acceleration of 12 ft/s2 due to the action of a resultant force applied for 6.5 s. Determine the work of the resultant force in ft. lbf and in Btu. m = 80 lb. Acceleration = 12.0 ft/s2. time of application = 6.5s Terminal velocity V2 = 12.0 x 6.5 = 78 ft/s Hence, Work = change in KE = ½ (80) (78.0)2 lbm ft2/s2 = (40) (78)(78) / 32.2 ft. lbf = ft.lbf = / 778 Btu = Btu.

Problem 2.27 A gas is compressed from V1 = 0.3 m3, p1 = 1 bar to V2 = 0.1 m3, p2 = 3 bar. Pressure and volume are related linearly during this process. For the gas, find the work in kJ. First, the gas is compressed, hence work is negative. Second, let us first derive an expression for expansion/compression work before we solve the problem. p-V relation is linear: dV/dp = - 10; p = 4 bars, when V tends to zero (although not feasible); Hence p = 4 – 10 V

Integration of pdV

Problem 2.33 Carbon monoxide contained within a piston-cylinder assembly undergoes three processes in series: Process 1-2: Expansion from p1 = 5 bar, V1 = 0.2 m3, to V2 = 1 m3, during which the pressure volume relationship is pV = constant. Process 2-3: constant-volume heating from state 2 to state 3, where p3 = 5 bar. Process 3-1: constant pressure compression to the initial state. Sketch the process in series on p-V coordinates and evaluate work for each process in kJ.

Problem 2.33 Process 1-2: p V = constant. p2 = p1 V1 / V2 = (5 bars) (0.2m3)/(1m3) = 1 bar = 100 kPa Since pV = constant, Work = p1 V1 ln (V2/V1) Work = (500kPa) (0.2m3) ln (1/0.2) = 100 ln 5 = kJ Process 2-3: Constant volume heating. dV = 0. Hence work = 0. p3 = 5 bars Process 3-1: Constant pressure compression. Work = p3 (V1-V3) V3 = V2 = 1 m3; V1 = 0.2 m3; Work = -(500 kPa) (1 – 0.2) = -400 kJ

Problem 2.37 A 10-V battery supplies a constant current of 0.5 amp to a resistance for 30 min. a) determine the resistance, in ohms. b) for the battery, determine the amount of energy transfer by work in kJ. E = 10 V; i = 0.5 amp. R = E / i = 10/0.5 = 20 ohms. Work = Rate of work done x time = E i x time = 10 x 0.5 x 30 x 60 = 9000 Watts .s = 9kJ

Homework # 2 Problems from text: 2.14, 2.25, 2.28, 2.30, 2.38

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