A 23.0 kg child plays on a swing having support ropes that are 2.10 m long. A friend pulls her back until the ropes are 41.0-degrees from the vertical and releases her from rest.
a.)What is the potential energy for the child just as she is released compared with the potential energy at the bottom of the swing?
b.) How fast will she be moving at the bottom of the swing?
c.) How much work does the tension in the ropes do as the child swings from the initial position to the bottom?
2. Relevant equations
Potential Energy (U) = mgr
Kinetic Energy (K) = 05*m*v^2
W(noncons force) = ΔE = E(initial) + W(nc) = E(final)
3. The attempt at a solution
a.) U(top)= mgr
Potential energy at the top = (mass of child)(gravity)(radius or length of ropes)
= 23kg * 9.8m/s^2 * 2.10m
U(top) = 473 J
b.) U(top) = KE(bottom) so, mgr = .5mv^2
Mass cancels as it is on both sides of the equation, so you’re left with: gr = .5v^2
2gr = v^2 then square-root the whole thing to yield 2gr^.5 = v
v = (2*9.8*2.10)^.5
v = 6.42 m/s
c.) I think nonconservative forces Work equation should come into play here:
W(nc) = ΔE
E(i) + W(nc) = E(f) or E(i) = E(f) + W(nc) <–not sure which way this should go. Also not sure how to go about solving for the Tension.
Thank you so much for any help.