Gravitational Potential Energy-Work (ΔU) Question

1. Jan 29, 2017

Arman777

1. The problem statement, all variables and given/known data
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 ?
2. Relevant equations
$(ΔU)=U_f-U_i$
$-ΔU=W$
$W=\frac 1 2 m((v_f)^2-(v_i)^2)$

3. The attempt at a solution

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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2. Jan 30, 2017

Staff: Mentor

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.

3. Jan 30, 2017

Arman777

I put A and B and H other then that its just like the book.It can go higher but I dont know where it can go max ? and if its I dont know the angle to find the "how high"?

4. Jan 30, 2017

Staff: Mentor

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).

5. Jan 30, 2017

Arman777

I solved thanks.Just it sounded awkward