1. Kinetic Energy m s F v E k = 1 / 2 mv 2 E k - Kinetic energy v - velocity m - mass

2. 1960 J E k = 1 / 2 mv 2 = 1 / 2 (.0042 kg)(965 m/s) 2 = 1955.5725 (1960 J) What is the kinetic energy of a 4.20 g bullet going 965 m/s? (units?)

3. 11 m/s E k = 1 / 2 mv 2 v =  (2 E k /m) =  (2(34 J)/(.563 kg)) = 11 m/s What speed must a.563 kg hammer move to store 34 J of energy?

4. 90. J initial:E k = 1 / 2 mv 2 = 1 / 2 (4.0 kg)(6.0 m/s) 2 = 72 J final:E k = 1 / 2 mv 2 = 1 / 2 (4.0 kg)(9.0 m/s) 2 = 162 J So the increase is 162 – 72 = 90. J increase A 4.0 kg shot is sped up from 6.0 m/s to 9.0 m/s. What is the change in kinetic energy?

5. 20.29 g E k = 1 / 2 mv 2 m = 2 E k /v 2 = 2(2.055 J)/(14.23 m/s) 2 =.020297… (20.29 g) A European swallow has 2.055 J of kinetic energy when it is flying at 14.23 m/s. What is its mass in grams?