Answer:
a. 139.748J
b.3.486m/s
c.zero
Explanation:
a. Given the mass of the child as 23.0kg, rope length is 2.1mand incline is 45°
Potential energy during release is calculated as:[tex]PE=mgh[/tex]
#Find vertical difference of when the swing is at rest (2.1m) and when the child is pulled back.
Find the height when the child is pulled back:
[tex]cos 45\textdegree=y/2.10\\\\y=1.48m[/tex]
#therefore,vertical difference is 2.1m-1.48m=0.62m
[tex]\therefore PE=mgh\\\ \ =23.0kg\times 9.8m/s^2\times 0.62m\\\ \ =139.748J[/tex]
#Hence the potential energy during release is 139.748J
b. From a, above, we have PE=139.748J, M=23.0kg.
At the bottom, all the PE will be transferred into KE. Potential energy is calculated as:
[tex]KE=0.5mv^2\\\\mv^2=2KE\\\\v=\sqrt{2KE/m}\\\\v=\sqrt{2\times 139.748J/23.0}\\\\v=3.486m/s[/tex]
#Hence the velocity at the bottom of the swing is 3.486m/s
c. Work is calculated as the product of force by distance.
From a, b above we have mass as 23.0kg .
-since the distance of the ropes remained constant the change in distance is zero:
[tex]W=mgd\\=23.0\times 9.8m/s^2\times 0\\=0[/tex]
Therefore the work in the ropes is 0,zero.
