William C. answered 11/14/23
Experienced Tutor Specializing in Chemistry, Math, and Physics
I modified my initial answer where I tried to work this as a momentum-impulse problem.
It makes more sense to treat it as a work-energy problem.
So we start off by setting work (force × distance) equal to
the loss of kinetic kinetic energy of the boxer's hand.
a)
Plug in m = 7.25 kg, v = 9.5 m/s, and xcomp = 7.25 cm = 0.0725 m
to get F = (7.25)(9.5)2/(0.0725)2 = 4512.5 N
b)
with xcomp = 2 cm = 0.02 m
F = (7.25)(9.5)2/(0.02) = 16,358 N