Kinetic and Potential Energy Problem Solving

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The discussion revolves around solving problems related to kinetic and potential energy involving a mass on a string. For the first question, the angle of the string with the vertical when the height is 0.1m is calculated to be approximately 25.84 degrees using trigonometric principles. The second question involves determining the tension in the string at that height, which requires considering the forces acting on the mass, including the need for centripetal acceleration. The third question similarly addresses the tension when the mass returns to its original position, emphasizing that tension must account for both gravitational force and centripetal force. Overall, the participants are exploring the relationships between energy, tension, and motion in this context.
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Homework Statement


Doing some homework with kinetic energy and potential energy. I came across 3 questions that I am struggling with. Basically we have a mass attached to the end of a string. The string is hanging straight down to start off these questions. The following questions assume that the string never has slack in it!

#1, If the length of the string is 1m, calculate the angle made by the string with the vertical when height=.1m?

#2, If the mass is .06kg, what is the tension on the string when height=.1m?

#3, What is the tension of the string when the ball falls back to its original position from a height .1m?

Homework Equations


Well, I know mgh=1/2mv2, the m will cancel out leaving us with, gh=1/2v2

The Attempt at a Solution


For #1, If I draw this out, we will have a shape that looks like a right triangle with the hypotenuse of 1m and the vertical side of 1m-height (.1)=.9m. Using SOHCAHTOA we should get an angle of 25.84 degrees.

For #2, I'm not sure about this one, I think I have to label all the forces acting on the mass and do something from there.

For #3,Shouldn't the tension of the string be the Net Force=ma? The only tension acting on the string in the initial position (which is at 0 degrees) should be just the .6kgx9.8?
 
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For #1, If I draw this out, we will have a shape that looks like a right triangle with the hypotenuse of 1m and the vertical side of 1m-height (.1)=.9m. Using SOHCAHTOA we should get an angle of 25.84 degrees.

Yup.

For #2, I'm not sure about this one, I think I have to label all the forces acting on the mass and do something from there.

I'm not sure either. If the mass is moving at h=0.1 m, the tension will be higher, because it has to provide the required amount of centripetal acceleration. However, I think the question assumes that the mass starts off at this position, so it isn't moving at this time.

For #3,Shouldn't the tension of the string be the Net Force=ma? The only tension acting on the string in the initial position (which is at 0 degrees) should be just the .6kgx9.8?

Tension also has to provide the centripetal acceleration, so it's going to be more than 0.6kg*9.8 m/s^2.
 
So I'm guessing for #2, It would be something along the lines of T1+T2=?... most likely the value of centripetal force.

Same for #3 I guess too.
 
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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