Solution: The forces acting on the mass are shown in figure. Apply Newtons second law of motion we have (a)Apply Newton's second law of motion mg-f= ma, we have dt That is =mg- (b)When the mass reaches its terminal speed, the acceleration does not change mg-f=ma=0, thus mg -Bv,=m-=0,that is (c)according to the answer in(b), we know 0. So mg -Bv =0 The terminal speed is (d) If the mass is dropped from rest, then dt d(mg-Bv, B ng-Bi >In(mg-B,)-In(mg)=-BILZ →加、B,)=-B B,1%z fern (1Solution: The forces acting on the mass are shown in figure. Apply Newton’s second law of motion, we have (a) Apply Newton’s second law of motion mg − f = ma , we have t v mg v ma m y y y d d − β = = That is y y mg v t v m = − β d d (b) When the mass reaches its terminal speed, the acceleration does not change, so mg − f = ma = 0 , thus 0 d d − = = t v mg v m y β y , that is 0 d d = t vy . (c) according to the answer in (b), we know 0 d d = t vy . So mg − βvy = 0 . The terminal speed is β mg vterm = . (d) If the mass is dropped from rest, then (1 ) (1 ) 1 ln(1 ) ln( - ) ln( ) d - d( - ) dt d d d vy 0 0 m t term m t y m t y y y t y y y y y y e v e mg v v e mg m t v mg m t mg v mg t mg v m mg v mg v m v mg v t v m β β β β β β β β β β β β β β − − − ⇒ = − = − ⇒ − = ⇒ − = − ⇒ − = − ⇒ = − = − = − ⇒ ∫ ∫ mg v f v j ˆ