Gravitational Potential Energy-Work (ΔU) Question

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Homework Help Overview

The problem involves a ball attached to a string that swings from a fixed point, with a peg affecting its motion. The discussion focuses on calculating the speed of the ball at its lowest and highest points, utilizing concepts from gravitational potential energy and work.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply energy conservation principles to find the speeds at different points in the swing. Some participants question the identification of the highest point and the implications of the string's length on the ball's motion.

Discussion Status

Participants are actively discussing the setup of the problem, with some providing clarifications regarding the positions of points A, B, and the highest point. There is recognition that the original poster's calculations may not align with the book's answer, prompting further exploration of the assumptions made.

Contextual Notes

There is uncertainty regarding the exact height the ball can reach after passing point B, as well as the angles involved in determining this maximum height. The discussion also hints at the need to consider the tautness of the string throughout the motion.

Arman777
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Homework Statement


The string (in the pic) is L=120 cm long,has a ball attached to one end,and is fixed as its other end.The distance d from the fixed point end to a fixed peg at point P is 75.0cm.When the initally stationary ball is released with the string horizontal as shown,it will swing along the dashed arc.Whats its speed when it reaches (a) its lowest point and (b) its higest point after the string catches on the peg ?

Homework Equations


##(ΔU)=U_f-U_i##
##-ΔU=W##
##W=\frac 1 2 m((v_f)^2-(v_i)^2)##

The Attempt at a Solution



[/B] İts lowest point is in A.Using;
##-ΔU=W##
##W=\frac 1 2 m((v_f)^2-(v_i)^2)##
we get

##-mgH+mg(H+d+r)=\frac 1 2m(v_f)^2-0##
##2gL=(v_f)^2##
##2.9.8\frac {m} {s^2}.1.2m=(v_f)^2##
##v_f=4.85 \frac m s##
which its speed in the lowest point.And answer for part (a)

For part (b),The object has inital speed ##v_i=4.85 \frac m s## and goes point B.
same equations;

##-mg(H+r)+mg(H)=\frac 1 2 m((v_f)^2-(v_i)^2)##
##\frac 1 2m(v_i)^2-mgr=\frac 1 2 m(v_f)^2##
If we multiply by 2 and divide m we get
##(v_i)^2-2gr=(v_f)^2##
##((4.85 \frac m s)^2)-(2.9.8\frac {m} {s^2}.0.45m)=(v_f)^2##
##23.52\frac {m^2} {s^2}-8.82\frac {m^2} {s^2}=(v_f)^2##
##v_f=3.83\frac m s##

which book says ##v_f=2.42\frac m s##
Where did I go wrong ?

Thanks
 

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Is the sketch from the book?

The highest point is not at B (as it is marked in the sketch). Your speed is correct for point B, but it will continue to go up beyond that point. The answer of the book takes that into account.
 
mfb said:
Is the sketch from the book?

The highest point is not at B (as it is marked in the sketch). Your speed is correct for point B, but it will continue to go up beyond that point. The answer of the book takes that into account.

I put A and B and H other then that its just like the book.It can go higher but I don't know where it can go max ? and if its I don't know the angle to find the "how high"?
 
Arman777 said:
but I don't know where it can go max ?
Think of the string length. What is the distance of both A and B to P? What will be the distance of the highest point to P? You can assume that the string will stay taut (you can also prove this, but that is beyond the scope of this question).
 
mfb said:
Think of the string length. What is the distance of both A and B to P? What will be the distance of the highest point to P? You can assume that the string will stay taut (you can also prove this, but that is beyond the scope of this question).

I solved thanks.Just it sounded awkward
 

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