Find displacement of marble launched from a spring (Energy), help please

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A marble is launched horizontally from a spring with a constant of 12 N/m, compressed by 4.0 cm, and falls 93.0 cm to the floor. The initial energy calculation yields 0.0096 J, leading to a velocity of 2.31 m/s, which was mistakenly identified as the final velocity. The correct approach requires taking the square root to find the actual velocity, which affects the horizontal distance traveled. The calculated distance of 1.00716 m differs from the textbook's answer of 0.66 m, indicating an error in the velocity determination. Clarification on the velocity calculation is essential for resolving the discrepancy.
mickeymouseho
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1. A horizontal spring, of constant 12N/m, is mounted at the edge of a lab bench to shoot marbles at targets on the floor 93.0 cm below. A marble of mass 8.3 x 10^-3 kg is shot from the spring, which is initially compressed a distance of 4.0 cm. How far does the marble travel horizontally before hitting the floor?



2. Relevant equations
E = 0.5kx^2
E = 0.5mv^2
delta d = 0.5(acceleration)(delta time)^2
dx = vx x t



The Attempt at a Solution



E = 0.5kx^2
= 0.5(12N/m)(0.04m)^2
= 0.0096J

E = 0.5mv^2
0.0096J = 0.5(8.3x10^-3kg)(v)^2
Square root of [0.0096J / (0.5x8.3x10^-3kg)] = v
2.31 m/s = v

delta d = 0.5at^2
t = square root of [ (2 times delta d) / a] *accleration = gravity, 9.8m/s
t = square root of [ (2 x 0.93m) / (9.8m/s^2)]
t = 0.436s

dx = vx x t
= 2.31 m/s x 0.436s
= 1.00716m

I've got 1.00716m as the answer. However, the textbook says the answer is 0.66m. I'm guessing I did something wrong at finding the velocity. I can't seem to figure this problem out. Please help. Thanks in advance! :smile:
 
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mickeymouseho said:
1.A horizontal spring, of constant 12N/m, is mounted at the edge of a lab bench to shoot marbles at targets on the floor 93.0 cm below. A marble of mass 8.3 x 10^-3 kg is shot from the spring, which is initially compressed a distance of 4.0 cm. How far does the marble travel horizontally before hitting the floor?

2. Homework Equations
E = 0.5kx^2
E = 0.5mv^2
delta d = 0.5(acceleration)(delta time)^2
dx = vx x t

3. The Attempt at a Solution
E = 0.5kx^2
= 0.5(12N/m)(0.04m)^2
= 0.0096J

E = 0.5mv^2

0.0096J = 0.5(8.3x10^-3kg)(v)^2
Square root of [0.0096J / (0.5x8.3x10^-3kg)] = v
2.31 m/s = v


I've got 1.00716m as the answer. However, the textbook says the answer is 0.66m. I'm guessing I did something wrong at finding the velocity. I can't seem to figure this problem out. Please help. Thanks in advance!

Welcome to PF.

2.31 is V2 not V
 
LowlyPion said:
Welcome to PF.

2.31 is V2 not V

Thank you for the welcoming! :biggrin:

And yes, I am a clutz, for forgetting to square root... :blushing::redface: Gosh this is embarrasing...

Thanks so much for the help!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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