a) Using conservation of energy, we can find the child's speed at the lowest position:
Initial energy = Potential energy at highest point
1/2(mv^2) + mgh = mgh
1/2(v^2) = g(2 - 2cos37)
v^2 = 19.6(2 - 2cos37)
v = 4.19 m/s
Therefore, the child's speed at the lowest position is 4.19 m/s.
b) To find the energy loss due to friction, we can use the conservation of energy again:
Initial energy = Final energy
1/2(mv^2) + mgh = 1/2(mv_l^2) + mgh_l + E_loss
where v_l is the speed at the lowest position and h_l is the height at the lowest position.
Solving for E_loss:
E_loss = 1/2(mv^2 - mv_l^2) - mg(h - h_l)
E_loss = 1/2(30)(4.19^2 - 2.2^2) - 9.8(2 - 2cos37)
E_loss = 39.7 J
Therefore, the energy loss due to friction is 39.7 J.
A 30 kg child on swing 2m long is released from rest
when the swing supports make an angle of 37 with the
vertical
a) neglecting friction, find the child's speed at lowest position
b) if the speed of the lowest position is 2.2m/s, what is the energy loss
due to friction?
1 answer