Question

A 20.0 kg child is on a swing that hangs from 3.00 m-long chains. What is her speed v(i) at the bottom of the arc if she swings out to a 45.0 degree angle before reversing direction? The answer is in m/s.

So far I came up with 6.45 and 18 which are both incorrect. Please help

Answer 1

If she swings out to 45° then the vertical component of the swing chains will be 3cos45° so the amount she will have risen will be 3 - 3cos45° which is about 0.88m

Now from that you can work out her potential energy which will be mass x height = 20 x 0.88 x g = 172.48J

Assuming no losses that means that her kinetic energy at the bottom of the swing will be the same, so from 172.48 = 1/2mv^2

v^2 = 17.248 so v= 4.15m/s

Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat.

SHOW ANSWER

Answer 2

This Site Might Help You.

RE:

A 20.0 kg child is on a swing that hangs from 3.00 m -long chains?

A 20.0 kg child is on a swing that hangs from 3.00 m-long chains. What is her speed v(i) at the bottom of the arc if she swings out to a 45.0 degree angle before reversing direction? The answer is in m/s.

So far I came up with 6.45 and 18 which are both incorrect. Please help

Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat.

SHOW ANSWER