DATA TABLE Distance mass was lifted (m) 0.5 Voltage rating of motor (V) 15 Electrical energy Mechanical Run Load lifted used energy Efficiency (9) ( J ) output (%) (J ) 10 0.496 0.049 9.88% 2 20 1.209 0.098 8.11% 3 30 1.439 0.147 10.24% 4 40 1.493 0.196 13.13% 5 50 2.2112 0.245 11.08% 6 60 2.269 0.294 12.96% 7 70 2.955 0.343 11.61%ANALYSIS 1. For each experimental run, calculate the increase in gravitational potential energy of the mass in joules. The increase in gravitational potential energy is equal to the mechanical energy output of the motor. Record the values in the data table. (10/1000)(9.81)(0.5) = 0.04905 (20/1000)(9.81)(0.5) = 0.0981 (30/1000)(9.81)(0.5) = 0.14715 (40/1000)(9.81)(0.5) = 0.1962 (50/1000)(9.81)(0.5) = 0.24525 (60/1000)(9.81)(0.5) = 0.2943 (70/1000)(9.81)(0.5) = 0.34335 2. For each run, calculate the efficiency of the motor; that is, what percentage of the electrical energy into the motor was converted to gravitational potential energy? Record your answer in the data table. (0.049/0.496) x 100 = 9.88% (0.0981/1.209) x 100 = 8.11% (0.147/1.436) x 100 = 10.24% (0.196/1.493) x 100 = 13.13% (0.245/2.2112) x 100 = 11.08% (0.294/2.269) x 100 = 12.96% (0.343/2.995) x 100 = 11.61% 3. For which load was the motor most efficient? The load which was most efficient was the 40g load due to its equivalent to 13.13% efficiency . What happened to the remainder of the electrical energy that went to the motor? The remainder of the electrical energy would be converted to electromagnetic force to spin and dissipated as heat, dispersed to the surroundings attributed to loss of energy of the undesired heat in electronic parts, friction loss, and others. 5. How does the typical efficiency of your motor compare to the fraction of electrical energy converted to work that you guessed in the Preliminary Questions