A 31.0 kg child on a 3.00 m long swing is released from rest?

Question

when the ropes of the the swing make an angle of 28.0° with the vertical

(a) Neglecting friction, find the child's speed at the lowest position.____m/s

(b) If the speed of the child at the lowest position is 2.30 m/s, what is the mechanical energy lost due to friction?____J.

hey guys I keep getting the wrong answer. Can you show me how to get to the answer? Thanks

The second answer is wrong according to my book.

Answer 1

draw the swing like at the link to get all data u need
http://img179.imageshack.us/img179/8470/swinget4.jpg
(I hope u know about this geometry)
then
(a)
potential energy = keinetic energy
mgh = 1/2 mv²
0.35 * 9.8 = 0.5 v²
v = √(2 *9.8*0.35)
v = 2.62 m/s (without friction)

(b)
Energy without friction = 1/2 mv² =
1/2 * 31 * 2.62² = 106.68 J

Energy with friction = 1/2 mv² =
1/2 * 31 * 2.3² = 81.995J

Energy lost due to friction = 106.68 - 81.995 = 24.685J

I hope this helps

Answer 2

1. draw a picture of the situation
2. find the inital height of the child in the swing- all of his energy is potential energy--- mgh=U
3. at the lowest point, all of the energy is kinetic---- 0.5*m*v^2=mgh
3. the velocity from (2) is if there is no friction, so the difference in energy is the loss due to friction

Answer 3

a) swing length = 3 m
swing angle = 28 deg
When child is rest, potential energy = mgh
When child at lowest position, kinetic energy = 1/2mv^2
Since there is no fiction, all potential energy convert to kinetic energy.
mgh = 1/2mv^2
gh = 1/2 v^2
(9.81)(3 - 3 cos 28) = 1/2 v^2
v^2 = 6.8897
v = 2.6248 m/s

b) Child weight = 31kg
Due to fiction, speed of child at lowest position = 2.30m/s
mgh = 1/2mv^2 + energy loss
(31)(9.81)(3 - 3 cos 28) = 1/2(31)(2.3)^2 + energy loss
106.79 = 82 + energy loss
energy loss = 24.79 J