Uncovering the Mass of an Object on Planet X

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Homework Help Overview

The problem involves determining the weight of a rock on Planet X after it is thrown upward. The rock has a mass of 5.24 kg and is thrown at an initial velocity of 13.0 m/s, returning to the same level after 1.51 seconds. The context suggests a focus on gravitational acceleration in a non-Earth environment.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss various equations of motion to find acceleration, including the use of displacement and time. There are attempts to clarify the relationship between initial velocity, final velocity, and acceleration. Some participants question the assumptions made about the motion and the calculations involved in determining weight.

Discussion Status

The discussion is ongoing, with multiple approaches being explored to calculate acceleration and weight. Some participants have provided alternative equations and methods, while others are questioning the correctness of the initial calculations. There is no explicit consensus on the correct approach yet.

Contextual Notes

Participants note that the weight calculation depends on the acceleration due to gravity on Planet X, which is not provided. There is also mention of the online homework system indicating incorrect answers, prompting further investigation into the calculations.

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Homework Statement


When venturing forth on Planet X, you throw a 5.24kg rock upward at 13.0 m/s and find that it returns to the same level 1.51 s later.

What does the rock weigh on Planet X?

Homework Equations



W=m*a

The Attempt at a Solution



To find the acceleration i good the 13.0m/s / 1.51s = 8.61 m/s^2. W=5.24kg * 8.61 m/s^2 = 45.1 N

The online homework keeps telling me that my answer is wrong. What am I doing wrong? Thank you for the help!
 
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Try using this equation:
x = (x-initial) + (V-initial)(t) + (1/2)(a)(t^2)
(remember that in your case the change in x = 0, which means x - (x-initial) = 0)

Someone correct me if I am wrong :0
 
With this formula could be easier: Vf=Vo+at.
Vo is your initial velocity, Vf is the final velocity that for you is the velocity at highest point the stone reaches (which is 0), and the time for you is 1.52/2 because you're just dealing with half of the experiment.
 
You can use: [tex]s=ut+\frac{1}{2}at^{2}[/tex] for the entire 'experiment'.

[tex]s=displacement[/tex]
[tex]u=initial (starting) velocity[/tex]
[tex]t=time[/tex]
[tex]a=acceleration[/tex]

Note: You are trying to find the acceleration so the weight formula can be used later.

Therefore, by rearrangement, [tex]a=\frac{2(s-ut)}{t^{2}}[/tex]

Since the rock is launched at 13ms-1, this is the initial velocity (u)
The time taken for the rock to come back from where it was launched, t=1.51 seconds
This seems to be all the information given that is relevant to this equation. But it is also known that the displacement (s) will be 0 when the rock reaches back to where it left.

Hence, [tex]a=\frac{2(0-13(1.51))}{(1.51)^{2}}[/tex]

[tex]a \approx -17.22 ms^{-2}[/tex] (note: negative acceleration means the acceleration is acting opposite to the rock launch. i.e. downwards towards the planet)

Now since the acceleration has been found, the weight of the rock can be found using:
[tex]F=ma[/tex]

F= force (Newtons) / weight of object
m = mass = 5.24kg
a = acceleration = -17.22 ms-2
Therefore, [tex]F=(5.24)(17.22)[/tex]

[tex]F\approx 90kg[/tex]
 

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