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Betty, with a mass of 46.8 kilograms, sits on a playground swing that hangs 0.417 meters above the ground. Brian pulls the swing back and releases it when the seat is 1.22 meters above the ground. If 179 joules of energy is converted to thermal energy, how fast does Betty move through the lowest position? Include units in your answer. Answer must be in 3 significant digits.

User Ibu
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In order to determine the sped of Betty at the lowest position, proceed as follow:

The potential energy for a height of 1.22m is:


\begin{gathered} U_1=m\cdot g\cdot h_1 \\ U_1=(46.8kg)(9.8(m)/(s^2))(1.22m) \\ U_1\approx559.54J \end{gathered}

The potential energy at the lowest point is:


\begin{gathered} U_2=m\cdot g\cdot h_2 \\ U_2=(46.8kg)(9.8(m)/(s^2))(0.417m) \\ U_2\approx191.25J \end{gathered}

Consider that at the starting point all the mechanical energy is potential energy (559.54J). Moreover, when Betty moves through the lowest point, part of the mechanical energy becomes thermal energy and kinetic energy.

Take into account that the for the lowest point, the sum of thermal energy, kinetic energy and potential energy must be equal to the total mechanical energy:

U2 + K + Th = 559.54J

where K is the kinetic energy and Th is the thermal energy. Consider that

Th = 179J

U2 = 191.25J

Solve the equation for K and replace the values of the other contributions to the total energy:

K = 559.54J - U2 - Th

K = 559.54J - 191.25J - 179J

K = 189.29J

Next, take into account that the kinetic energy is given by:


K=(1)/(2)mv^2

solve the previous equation for v and replace the values of K and m:


\begin{gathered} v=\sqrt[\square]{(2K)/(m)} \\ v=\sqrt[]{\frac{2(189.29J)}{46.8\operatorname{kg}}} \\ v\approx2.84(m)/(s) \end{gathered}

Hence, Betty moves with a speed of approximately 2.84m/s through the lowest point.

User LabRat
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