A baseball pitcher throws a 0.14kg baseball with a final speed of 20m/s. a. What is the kinetic energy of the baseball when it is at the final speed? b. If the baseball was initially at rest, how much work was done to give it this kinetic energy?

Solution
Given: m = 0.14kg
vi = 0m/s
vf = 20m/s

Unknowns: a. kinetic energy, Ek b. work, W

Basic equations: a. Ek = ½mv2 b. W = ΔEk

Solution: a. Ek = ½mv2 = ½(0.14 kg)(20 m/s)2
= 28 kg * m2/s2
= 28 J b. Work done is equal to the change in kinetic energy.
W = ΔEk = Ekf - Eki
= 28 J – 0 J
W = 28J

a.What is the kinetic energy of the baseball when it is at the final speed?b.If the baseball was initially at rest, how much work was done to give it this kinetic energy?SolutionGiven: m = 0.14kg

vi = 0m/s

vf = 20m/s

Unknowns:

a.kinetic energy, Ekb.work, WBasic equations:

a.Ek = ½mv2b.W = ΔEkSolution:

a.Ek = ½mv2 = ½(0.14 kg)(20 m/s)2= 28 kg * m2/s2

= 28 J

b.Work done is equal to the change in kinetic energy.W = ΔEk = Ekf - Eki

= 28 J – 0 J

W = 28J