Question

A 0.18kg apple falls from a tree to the ground, 4.0 {\rm m} below. Ignore air resistance. Take ground level to be y = 0. Part 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. Part 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 Part 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

Answer 1

At h = 4.0
K = 0.5 (0.18) (0)² = 0
U = (0.18) (9.81) (4.0) = 7.06 J
E = K + U
E = 0 + 7.06 = 7.06 K

At h = 3.0
K = 0.5 (0.18) (2)(9.81)(1.0) = 1.76 J
U = U = (0.18) (9.81) (3.0) = 5.30 J
E = 1.76 + 5.30 = 7.06 J

At h = 2.0
K = 0.5 (0.18) (2)(9.81)(2.0) = 3.53 J
U = U = (0.18) (9.81) (2.0) = 3.53 J
E = 3.53 + 3.53 = 7.06 J

Answer 2

A) At h=4.0 m

At h=4.0 m, the kinetic energy of the ball is zero, because its velocity is 0, so

`K=(1)/(2)mv^2=(1)/(2)(0.18 kg)(0)^2=0`

The gravitational potential energy is instead:

`U=mgh=(0.18 kg)(9.8 m/s^2)(4.0 m)=7.06 J`

So, the total mechanical energy is

`E=K+U=0+7.06 J=7.06 J`

B) At h=3.0 m

The gravitational potential energy is now:

`U=mgh=(0.18 kg)(9.8 m/s^2)(3.0 m)=5.29 J`

Since air resistance is negligible, the total mechanical energy is conserved, so it is still

`E=7.06 J`

And so we can find the kinetic energy as follows:

`K=E-U=7.06 J-5.29 J=1.77 J`

C) At h=2.0 m

The gravitational potential energy is now:

`U=mgh=(0.18 kg)(9.8 m/s^2)(2.0 m)=3.53 J`

Since air resistance is negligible, the total mechanical energy is conserved, so it is still

`E=7.06 J`

And so we can find the kinetic energy as before:

`K=E-U=7.06 J-3.53 J=3.53 J`