✔ 最佳答案
(1) m=5(kg), Vo=8(m/s), Vt=0, L=3(m), Q=30(deg)(a) ΔKE=m(Vt^2-Vo^2)/2=-5*64/2=-160(J)
(b) ΔPE=mg*Δh=mg*L*sinQ=5*9.8*3*1/2=15*4.9=73.5(J)
(c) 0=ΔKE+ΔPE+Ff*L=-160+73.5+Ff*3Ff=(160-73.5)/3=28.833(N)
(d) fk=Ff/mg*cosQ=28.833/5*9.8*cos30=0.679
2014-01-16 20:08:01 補充：
(2) m1=8kg, m2=10kg, a=1m/s^2, r=0.25m, Q1=30, Q2=60
(a)
F1-m1*g*sin30=m1*a =>
F1=m1*(a+g/2)
=8*(1+4.9)
=47.2(N)
m2*g**sin60-F2=m2*a =>
F2=m2*(g*0.866-a)
=10*(9.8*0.866-1)
=74.868(N)
2014-01-16 20:08:24 補充：
(b) T=?, I=?
T=(F2-F1)*r
=(74.868-47.2)*0.25
=6.917(N.m)
T=I*α=I*a/r =>
I=T*r/a
=6.917*0.25/1
=1.73(kg.m^2)
