5) suppose in the previous question when timmy is sitting at rest he has 40j of GPE then his dad geves timmy a push when his dad lets go Timmy has 60j of GPE and 12J of KE
a) befor his dad pushed him GPE= 40J Ke= 0J TE=40J
b) immedialtey after his dad pushes him GPE= 60 KE= 120 TE=180
C) How much work did timmys dad do when he pushed Timmy? 180J-40J=140J
D) At the top of Timmys path
GPE=180j KE=oJ TE= 180j
E) He swing back down through the bottom of his path
GPE=oj KE=180j TE=180j
F) Timmy jumps off the swing just befor he lands
GPE=oj KE=180j TE=180j
G) timmy hit the ground
GPE=oj KE=oj TE=oj
H) how much work does the ground do on (Didnt answer help please with letter h)
Nuts to these:
immedialtey after his dad pushes him GPE= 60 KE= 120 TE=180
C) How much work did timmys dad do when he pushed Timmy? 180J-40J=140J
nuts to these:
D) At the top of Timmys path
GPE=180j KE=oJ TE= 180j
E) He swing back down through the bottom of his path
GPE=oj KE=180j TE=180j
F) Timmy jumps off the swing just befor he lands
GPE=oj KE=180j TE=180j
H. at jumping off, he has zero gpe, but he has KE. The ground must have done that work to make the final KE zero.
After DAd swims him, dad must have done (60-40)J increase in GPE+ an additional 12j for the KE.
To calculate the work done by the ground on Timmy, you can use the work-energy principle, which states that the work done on an object is equal to the change in its total mechanical energy. In this case, as Timmy hits the ground, his total mechanical energy reduces to zero, so the work done by the ground will be equal to the initial total mechanical energy.
Given that the initial total mechanical energy (TE) is 180J, the work done by the ground will also be 180J.