Question

please answer 2.22, 2.38, and 2.44

2.4 A body of mass of 10 kg is placed 3 m above a plane. How much work was done to lift the body above the plane? 294.3 J 2.8 A body weighs 25 lb. If it is moving at 10 ft./s, evaluate its kinetic energy 38.86 ft.Ibf 2.22 A spring is compressed 5 in. from its equilibrium position. If the spring modulus k is 10 lbf /in. of deflection, how much work was done in foot-pounds to deflect the spring? 10.4 ft.Ibf 2.38 A gas expands according to the equation pv = 1000, where p is the pressure in kPa and v is the specific volume in ml/kg. If the gas pressure drops from 1000 to 500 kPa, how much work was done by the gas? 693.1 kJ/kg 2.44 A gas expands according to the equation p = -250V + 600, where Vis in mand p is in kPa. If the initial pressure is 575 kPa, the initial volume is 0.1 m, and the final volume is 0.4 m, determine the work done igre

Answer 1

* Problem 2.22 : Compressed length, x = 5 inch Spring modulus, k = 10 Lbb/in work done to deflect the spring = kx = 1 x 10 x 52 lbf-in = 125 Lbf in So, work done = 125 ft-lbf = 10.417 ft Ubb * Problem 8.38 : Given, PV = 1000 P = 1000 kPa P2 = 500kPa mo at constant temperature we can see that. PV=Constant So, Gas expand with isothermal process so, P. Vi = P2V2 = constant = 1000 In case of isothermal process work done = PV, Lolo) 1000X 11000 1000X In 2 Work done, wl= 693. 147 KIlig * Problem 2.44 Given, P=-250 V + 600 P = 575 kPa V, = 0.1 m2 V2 = 0.4m3 2 We know that , coork done -Spor so, ws Seasov +600) av

- Integrating with respect to de volume 70.4 w=-aso ng t + coox .- - 350x og? +602x0.4 + 35 04.04?cooxos so, w: -18.75 +180 = 161.25 kilky Area under P.V diagram gives work done work done = (Ve-) ( P2+P.) - - - e x(0,4-0.1) (575 + 500] -27 = 161.25 kJ / 19 V Rg = -250X0.4+600 - 500 kPa