Gravitational Potential Energy-Work (ΔU) Question

AI Thread Summary
The discussion revolves around calculating the speed of a ball attached to a string as it swings from a horizontal position. For part (a), the speed at the lowest point is determined to be 4.85 m/s using energy conservation principles. In part (b), the initial speed is correctly calculated, but the highest point reached by the ball is misidentified, leading to a discrepancy with the book's answer of 2.42 m/s. Participants clarify that the ball continues to ascend beyond the initial point B, and the maximum height must be assessed based on the string's length. The original poster resolves their confusion after receiving guidance on the calculations and concepts involved.
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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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