Two masses m1 = 5.40 kg and m2 which has a mass 80.0% that of m1, are attached to a cord of negligible mass which passes over a frictionless pulley also of negligible mass. if m1 and m2 start from rest, after they have each traveled a distance h = 2.00 m, use energy content to determine the following. (a) speed v of the masses

2.09 m/s OK, you have 2 masses.. m1 = 5.40 kg and m2 is 80% of that or 0.80 * 5.40 = 4.32 kg. With the setup, The total mass of the system is 5.40 kg + 4.32 kg = 9.72 kg The kinetic gained from the falling mass is 5.40 kg * 2 m * 9.8 m/s^2 = 105.84 kg m^2/s^2 The kinetic energy lost from the rising mass is 4.32 kg * 2 m * 9.8 m/s^2 = 84.672 kg m^2/s^2 So the total kinetic energy gained by the system is 105.84 kg m^2/s^2 - 84.672 kg m^2/s^2 = 21.168 kg m^2/s^2 Now this energy has been working on a system with a total mass of 9.72 kg, so 21.168 kg m^2/s^2 / 9.72 kg = 2.177777778 m^2/s^2 over a distance of 2m, so 2.177777778 m^2/s^2 / 2 m = 1.088888889 m/s^2 So the effective acceleration your system is undergoing is 1.089 m/s^2 Now to calculate the speed. Distance under constant acceleration is d = 0.5 A T^2, so plugging in the known values and solving for T gives: 2 m= 0.5 * 1.089 m/s^2 T^2 2 m = 0.5445 m/s^2 T^2 3.673094582 s^2 = T^2 1.92 s = T Multiply the time by the acceleration gives: 1.92 s * 1.089 m/s^2 = 2.09 m/s Note, the acceleration of 1.0888889 m/s^2 can be easily obtained using the relative masses of the two objects. We have an object of mass 1 and a second object of mass 0.8 for a total mass of 1.8, this mass is being accelerated by the difference of 0.2, so the effective acceleration is 0.2/1.8 = 0.111111 times that of the local gravity, which is 9.8 m/s^2. And 0.11111 * 9.8 = 1.088889 m/s^2.