All Questions Topic List
Others Questions
Previous in All Question Next in All Question
Previous in Others Next in Others
Question Number 16682 by tawa tawa last updated on 25/Jun/17
Answered by sandy_suhendra last updated on 25/Jun/17
F=0.2NΔx=1mm=10−3mk=FΔx=0.210−3=200N/ma)Δx=30mm=0.03mPEthespring=12k.Δx2=12×200×0.032=0.09Jbi)KEtheball=PEthespring12mv2=0.0912×0.025×v2=0.09v=2.7m/sbii)A=pointatthespringB=pointatthetopoftheloopKEA+PEA=KEB+PEB12m.vA2+m.g.hA=12m.vB2+m.g.hB12×2.72+0=12vB2+10×0.2vB2=3.29⇒vB=1.8m/sbiii)N=Fsf−w=m.vB2R−m.g=0.025×1.820.1−0.025×10=0.56N
Commented by tawa tawa last updated on 25/Jun/17
wow,Godblessyousir.
Terms of Service
Privacy Policy
Contact: info@tinkutara.com