Uncovering the Mass of an Object on Planet X

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SUMMARY

The weight of a 5.24 kg rock on Planet X is calculated using the formula W = m * a, where 'm' is the mass and 'a' is the acceleration due to gravity. The acceleration was determined to be approximately -17.22 m/s², leading to a weight of approximately 90 N. The calculations involved using kinematic equations to find the acceleration based on the rock's initial velocity of 13.0 m/s and the total time of 1.51 seconds for the rock to return to its original height. The negative sign indicates that the acceleration acts in the opposite direction of the initial launch.

PREREQUISITES
  • Understanding of Newton's second law of motion (W = m * a)
  • Familiarity with kinematic equations, specifically s = ut + (1/2)at²
  • Basic knowledge of acceleration and its implications in physics
  • Ability to perform algebraic manipulations to solve for unknowns
NEXT STEPS
  • Research the implications of negative acceleration in projectile motion
  • Learn more about kinematic equations and their applications in different scenarios
  • Explore the concept of weight in varying gravitational fields
  • Study the relationship between mass, weight, and acceleration due to gravity
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Students studying physics, educators teaching mechanics, and anyone interested in understanding the principles of motion and gravity on different celestial bodies.

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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: s=ut+\frac{1}{2}at^{2} for the entire 'experiment'.

s=displacement
u=initial (starting) velocity
t=time
a=acceleration

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

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

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, a=\frac{2(0-13(1.51))}{(1.51)^{2}}

a \approx -17.22 ms^{-2} (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:
F=ma

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

F\approx 90kg
 

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