Kinetic and Potential Energy Problem Solving

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SUMMARY

The discussion focuses on solving problems related to kinetic and potential energy involving a mass attached to a string. The first problem calculates the angle of the string with the vertical when the height is 0.1m, resulting in an angle of 25.84 degrees using trigonometric principles. The second problem addresses the tension in the string at this height, which requires consideration of centripetal acceleration. The third problem examines the tension when the mass returns to its original position, emphasizing that the tension must account for gravitational force and centripetal force.

PREREQUISITES
  • Understanding of kinetic energy and potential energy concepts
  • Familiarity with trigonometric functions, specifically SOHCAHTOA
  • Knowledge of Newton's laws of motion, particularly net force calculations
  • Basic principles of centripetal acceleration
NEXT STEPS
  • Study the derivation of the kinetic energy formula, KE = 1/2 mv²
  • Learn how to calculate tension in a string under varying conditions
  • Explore centripetal force calculations in circular motion scenarios
  • Investigate the relationship between potential energy and height in gravitational fields
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators seeking to enhance their understanding of energy concepts and problem-solving techniques in real-world applications.

Callen9
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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.
 

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