Aweightlifter works out at the gym each day. part of her routine is to lie on her back and lift a 41kg barbell straight up from chest height to full arm extension, a distance of 0.45m . part a: how much work does the weightlifter do to lift the barbell one time? part b: if the weightlifter does 23 repetitions a day, what total energy does she expend on lifting, assuming a typical efficiency for energy use by the body? part c: how many 500 calorie donuts can she eat a day to supply that energy?

A)180.8 J   B) 16,632 J  C) 8

Explanation:

A) The work done by the weightlifter, is just the product of the weight of

        the barbell (which is exactly equilibrated by the force applied by her,

        assuming the lifting happens at constant speed) times the distance

        travelled by the barbell:    

         W = F*d = 41 kg * 9.8 m/s * 0.45 m = 180. 8 J

B) If the weightlifter makes 23 repetitions a day, the total work done will    

       be 23 times the one needed for one try, as follows:

       W = 23* 180.8 J = 4,158 J

       Assuming a typical body efficiency of 25%, the energy expended by

       the body, must be 4 times the used for useful work, as follows:

       Ebody = W / 0.25 = 4,158 J / 0.25 = 16,632 J

C)    At first, we need to convert J to cal, as follows:

       1 cal = 4,186 J

       If a donut has 500 cal, this means that it will have 2,093 J.

       Dividing the total energy (in J) between the energy added by one  

       donut, and approximating to the nearest integer, we have:

       Ndonuts= 16,632 J / 2,093 J = 7.95 = 8 Donuts

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s