Question

A 5.36 kg object falls freely (ignore air resistance), after being dropped from rest. Determine the initial kinetic energy (in J), the final kinetic energy (in J), and the change in kinetic energy (in J) for the following.
(a) first meter of fallinitial kinetic energy? Jfinal kinetic energy? Jchange in kinetic energy? J(b) second meter of fallinitial kinetic energy? Jfinal kinetic energy? Jchange in kinetic energy

Answer 1

Answer: 52.53

Explanation:

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.