See Answer

Add Answer +20 Points

Community Answer

See all the answers with 1 Unlock

Get 4 Free Unlocks by registration

Get 4 Free Unlocks by registration

mass of box, m_(1)=7kgtheta=37^(@)length of inclined surface, d=4.2m coefficient of friction, mu=0.35 height of the bottom of incline, h=6mmass of sandbox, m_(2)=14kgfbd of boxf= fonce of frictionN= normal foncemg= force of gravity.equating components perpendicular to incline, N=mg cos theta^()equating components paralkel to incline, mg sin theta-f=ma-theta ( a is the arceleration)we know that=>mg sin theta-mu mg cos theta=ma=>a=g(sin theta-mu cos theta){:[=>a=9.8(sin 37^(@)-0.35 cos 37^(@))],[=9.8(0.6-0.31 xx0.798)],[=3.456ms^(-2)]:}As the block travels distance d=4.2m Hence, applying equation of motion,{:[v^(2)-o^(2)=2da],[=>v^(2)=2xx4.2 xx3.456=29.03],[=>v^(2)=5.388ms],[D_(v_(y))v_(v)]:}resolving int ... See the full answer