A 0.25kg apple falls from a tree to the ground, 4.0 m below. Ignore air resistan

Question

A 0.25kg apple falls from a tree to the ground, 4.0 m below. Ignore air resistance. Take ground level to be y=0.

A) Determine the apple's kinetic energy, K, the gravitational potential energy of the system,U, and the total mechanical energy of the system, E, when the apple's height above the ground is 4.0 m.

B) Determine the apple's kinetic energy, K, the gravitational potential energy of the system,U, and the total mechanical energy of the system, E, when the apple's height above the ground is 3.0 m.

C) Determine the apple's kinetic energy, K, the gravitational potential energy of the system,U, and the total mechanical energy of the system, E, when the apple's height above the ground is 2.0 m.

D) Determine the apple's kinetic energy, K, the gravitational potential energy of the system,U, and the total mechanical energy of the system, E, when the apple's height above the ground is 1.0 m.

E) Determine the apple's kinetic energy, K, the gravitational potential energy of the system,U, and the total mechanical energy of the system, E, when the apple's height above the ground is 0 m.

Explanation / Answer

gravitational potential energy U= mgh = E - K

kinetic energy K = 0.5mv2 = E - U

total mechanical energy of the system, E= K+U

Total energy of the system is conserved.. imples E = constant... same in all cases

a.) U = mgh = 0.25 * 9.8 * 4 = 9.8J

v=0 implies K = 0

E = 9.8J

b.) U = mgh = 0.25 * 9.8 * 3 = 7.35J

E = 9.8 J

K = 9.8 - 7.35 = 2.45 J

c.) U = 9.8 * 0.25 * 2 = 4.9J

E = 9.8J

K = 4.9J

d.) U = 0.25 * 9.8 * 1 = 2.45J

E =9.8J

K = 7.35 J

e.) U = 0

E = 9.8J

K = 9.8J

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