Physics, 24.06.2019 19:50 scasemere

You have landed on an unknown planet, newtonia, and want to know what objects will weigh there. you find that when a certain tool is pushed on a frictionless horizontal surface by a 12.2 n force, it moves 16.1 m in the first 2.40 s , starting from rest. you next observe that if you release this tool from rest at 11.0 m above the ground, it takes 2.58 s to reach the ground. what does the tool weigh on newtonia? what would it weigh on earth?

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Answer from: depaze6652

1. On Newtonia: 28.846N 2. On Earth:42.767N

Explanation:

The weight is the force with which a planet attracts an object with mass , due to the acceleration of gravity of said planet, and is given by the following equation:

W=m.g (1)

Knowing this, let's find the weight of an object on Newtonia and on Earth:

1. On Newtonia:

The weight in this planet is given by (1):

$W_{N}=m.g_{N}$ (2)

In addition, we are told that when a certain tool is pushed on a frictionless horizontal surface by a 12.2 N force, it moves 16.1 m in the first 2.40 s , starting from rest ( $X_{o}=0$ , $V_{o}=0$ and $t_{o}=0$ ).

By Newton's 2nd law of motion we know the force is given by:

F=m.a (3) Being the mass of the object and its acceleration.

Isolating : $m=\frac{F}{a}$ (4)

We already know F=12.2N , we need to find . In order to approach it, we will use the following equations related to the motion in one dimension:

$V=\frac{\Delta X}{\Delta t}=\frac{X-X_{o}}{t-t_{o}}$ (5)

$V=V_{o}+a.t$ (6)

From (5):

$V=\frac{16.1m-0}{2.40s-0}=6.708m/s$ (7)

Substituting (7) in (6):

6.708m/s=0+a(2.40s) (8)

Finding : $a=\frac{6.708m/s}{2.40s}=2.795m/s^{2}$ (9)

Substituting (9) in (4):

$m=\frac{12.2n}{2.795m/s^{2}}$ (10)

m=4.364kg (11)

On the other hand, we are told that if you release this tool from rest ( $V_{oy}=0$ and $t_{o}=0$ ) at y=11.0m above the ground, it takes t=2.58s to reach the ground.

In this case we will use the following equation:

$y=V_{oy}.t-\frac{1}{2}g_{N}.t^{2}$ (12)

$11m=0-\frac{1}{2}g_{N}(2.58s)^{2}$ (13)

Finding $g_{N}$ :

$g_{N}=-6.610m/s^{2}$ (14) Where the negative sign only indicates the acceleration is "downwards", to the center of the planet.

Substituting (14) in (1):

$W_{N}=(4.364kg)(6.610m/s^{2})$ (15)

$W_{N}=28.846N$ This is the weight of the tool on Newtonia.

2. On Earth:

In the case of our planet, we will use the same weight equation, knowing the acceleration due gravity on Earth is $g_{E}=9.8m/s^{2}$ :

$W_{E}=m.g_{E}$ (16)

$W_{E}=(4.364kg)(9.8mm/s^{2})$ (17)

Finally:

$W_{E}=42.767N$ This is the weight of the same tool on Earth.

Answer from: Quest

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hope this !

Answer from: Quest

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i have found an answer to your question but use it as information.

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You have landed on an unknown planet, newtonia, and want to know what objects will weigh there. you...