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Lebombo

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Hi, may I request help or quick review to ensure that I present an accurate understanding of this concept. I want to move begin learning about the free fall of one block onto another block, but would like to be sure that I understand the physics of work, force, and energy of one block. Thank you.

A 10 kg block is dropped from a height such that it free falls for 10 seconds before hitting the ground. When the block reaches the ground, it is driven 2 meters into the ground. Let acceleration of gravity be 10m/s^2 and all kinetic energy is transferred into work.

What is the net force of the block while it is in free fall?

How much work is done on the block by gravity while in free fall?

What is the average acceleration of the block as it's being slowed down by the ground?

What is the average net force of the block while it is being driven into the dirt?

What is the average force the block applies to the ground?

What is the average force the ground applies to block?

[/B]

[itex]x_{f}= \frac{1}{2}at^{2}+v_{0}t + x_{0}[/itex]

[itex] v_f = at + v_{0} [/itex]

[itex] v^{2}_{f} = v^{2}_{i}+ 2a(x_{f}-x_{0})[/itex]

F*d = [itex] \frac{1}{2}mv^{2}_{f} - \frac{1}{2}mv^{2}{i} [/itex]

[/B]

[itex] v_{f} = (10m/s^2)(10s)+ 0 = 100m/s [/itex]

[itex] x_{f} = \frac{1}{2}(10)(10)^{2}[/itex] = 500 meters

F = ma = (10kg)(10m/s^2) = 100N while in free fall

F * d = F * 500m = [itex] \frac{1}{2}(10kg)(100m/s)^{2} - \frac{1}{2}(10kg)(0m/s)^{2} = 50,000 J [/itex]

of kinetic energy. Therefore, gravity has done 50,000 Nm (or Joules) of work on the block during free fall for 500 meters.

Since the block continues to travel into the ground for 2 meters,

[itex] v_{f}^{2} = v_{i}^{2} + 2a(x_{f}-x_{0}) [/itex]

[itex] (0m/s)^{2} = (100m/s)^{2} + 2a(2m) [/itex]

[itex]a = 2500\frac{m}{s^{2}} [/itex]on average

Average Net Force of the block: F = ma = 10kg(2500m/s^2) = 25,000N

Since the ground brings the block back to rest, the ground also applies an average force of 25,000N on the block in the upward direction.

1. Homework Statement

1. Homework Statement

A 10 kg block is dropped from a height such that it free falls for 10 seconds before hitting the ground. When the block reaches the ground, it is driven 2 meters into the ground. Let acceleration of gravity be 10m/s^2 and all kinetic energy is transferred into work.

What is the net force of the block while it is in free fall?

How much work is done on the block by gravity while in free fall?

What is the average acceleration of the block as it's being slowed down by the ground?

What is the average net force of the block while it is being driven into the dirt?

What is the average force the block applies to the ground?

What is the average force the ground applies to block?

## Homework Equations

[/B]

[itex]x_{f}= \frac{1}{2}at^{2}+v_{0}t + x_{0}[/itex]

[itex] v_f = at + v_{0} [/itex]

[itex] v^{2}_{f} = v^{2}_{i}+ 2a(x_{f}-x_{0})[/itex]

F*d = [itex] \frac{1}{2}mv^{2}_{f} - \frac{1}{2}mv^{2}{i} [/itex]

## The Attempt at a Solution

[/B]

[itex] v_{f} = (10m/s^2)(10s)+ 0 = 100m/s [/itex]

[itex] x_{f} = \frac{1}{2}(10)(10)^{2}[/itex] = 500 meters

F = ma = (10kg)(10m/s^2) = 100N while in free fall

F * d = F * 500m = [itex] \frac{1}{2}(10kg)(100m/s)^{2} - \frac{1}{2}(10kg)(0m/s)^{2} = 50,000 J [/itex]

of kinetic energy. Therefore, gravity has done 50,000 Nm (or Joules) of work on the block during free fall for 500 meters.

Since the block continues to travel into the ground for 2 meters,

[itex] v_{f}^{2} = v_{i}^{2} + 2a(x_{f}-x_{0}) [/itex]

[itex] (0m/s)^{2} = (100m/s)^{2} + 2a(2m) [/itex]

[itex]a = 2500\frac{m}{s^{2}} [/itex]on average

Average Net Force of the block: F = ma = 10kg(2500m/s^2) = 25,000N

Since the ground brings the block back to rest, the ground also applies an average force of 25,000N on the block in the upward direction.

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