• Calculate the gain in energy it takes to raise her from the bottom of the arc to the 45 degrees angle using a diagram some trig and the equation Potential Energy, PE = mgh. At the top, her Kinetic Energy is zero. Then as she drops again, (nearly) all that Potential Energy must become Kinetic Energy again, so use the KE equation, KE=(1/2)mv^2 to find the speed.
  • KE high + PE high = KE low + PE low 0 + mgl(1-costheta)=1/2 mv^2 + 0 mass cancels out leaving gl(1-costheta)=1/2 v^2 thus Sq.rt of 2(g)(l)(1-cos theta) 2(9.8)(3)(1-cos45)= 4.1 m/s

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