Question

A 0.017-kg acorn falls from a position in an oak tree that is 18.5 meters above the ground.

Calculate the velocity of the acorn just before it reaches the ground (rounding your answer to the integer) and its kinetic energy when hitting the ground (rounding your answer to the nearest tenth).

Velocity _______________ m/s

Kinetic Energy _______________ J

Answer 1

The velocity of the acorn just before it reaches the ground is approximately 10 m/s, and its kinetic energy when hitting the ground is approximately 0.85 J.

Step-by-step explanation:

To calculate the velocity of the acorn just before it reaches the ground, we can use the principle of conservation of energy. The acorn starts with potential energy due to its height above the ground, and this potential energy is converted into kinetic energy when it falls. The equation for potential energy is PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. The equation for kinetic energy is KE = 0.5mv^2, where v is the velocity.

Using the given information, we can calculate the potential energy of the acorn: PE = (0.017 kg)(9.8 m/s^2)(18.5 m) = 3.068 J. Since all of the potential energy is converted into kinetic energy, we can equate the two equations: KE = 3.068 J.

Solving for the velocity, v, we get: 0.5(0.017 kg)v^2 = 3.068 J. Solving for v, we find that the velocity of the acorn just before it reaches the ground is approximately 10 m/s.

To calculate the kinetic energy when hitting the ground, we can use the equation: KE = 0.5mv^2. Plugging in the mass of the acorn and the velocity we just calculated, we get: KE = 0.5(0.017 kg)(10 m/s)^2 = 0.85 J. Therefore, the kinetic energy when hitting the ground is approximately 0.85 J.

Answer 2

Velocity= 19.05 m/s

Kinetic energy= 3.08 J

Step-by-step explanation:

The corn possess potential energy. Feom law of conservation of energy, all the potential eneegy will be converted to kinetic energy, with negligible friction.

Potential energy, PE= kinetic energy, KE

Potential energy is given by mgh while kinetic energy is given by
`0.5mv^(2)`

Equating

`mgh=0.5mv^(2)`

Making v the subject of the formula then

`v=\sqrt {2gh}`

Substituting 9.81 m/s2 for g and 18.5 m for h then the velocity will be

`v=\sqrt {2* 9.81* 18.5}=19.0517715711689\ m/s\\v\approx 19.05\ m/s`

To get the kinetic eneegy, since PE=KE AND KE is

`KE=0.5mv^(2)`

Substituting v with 19.05 and m with 0.017 kg we have

`0.5* 0.017* 19.05^(2)=3.08467125/ J\approx 3.08 J`

This can also be achieved by mgh since PE=KE hence 0.17*9.81*18.5=3.085245 J and the results are slightly different due to small energy losses and rounding off of values.