Question

A 5.36 kg object falls freely (ignore air resistance), after being dropped from rest. Determine the initial kinetic energy, the final kinetic energy, and the change in kinetic energy for the following.

(a) first meter of fall
initial kinetic energy 0 J
final kinetic energy ___ J

(b) second meter of fall
initial kinetic energy ___ J
final kinetic energy ___ J

Answer 1

m*g = 5.36k * 9.8N/kg = 52.53 N. = Wt.
of object.

a. KE = 0 J. = Initial KE
V^2 = Vo^2 + 2g*h
V^2 = 0 + 19.6*1 = 19.6
V = 4.43 m/s.
KE = 0.5m*V^2 = 2.68*4.43^2 = 52.59 J.

b. V^2 = Vo^2 + 2g*h
V^2 = 4.43^2 + 19.6*1 = 39.22
V = 6.26 m/s.
KEo = 2.68*4.43^2 = 52.59 J.
KE = 2.68*6.26^2 = 105 J.