(b)
\Weight of book box Fg = 209 N
\Gravitational acceleration g = 9.8 m/s
\Fg = mg
\m = 209/g
\m = 209/9.8
\m = 21.33 kg
\Shoved force Fs = 430 N
\Shoved angle with horizontal = 35º
\Friction coefficient µ = 0.75
\Book box moved distance s = 8 m
\Horizontal Shoved force Fsh = Fs(cos35) = 430*0.819
\Fsh = 352.24 N
\Vertical Shoved force Fsv = Fs(sin35) = 430*0.5736
\Fsv = 246.65 N
\Vertical component of forces Fn = Fsv + Fg
\Fn = Net force in vertical direction
\Fn = 246.65 + 209
\Fn = 455.65 N
\Force due to friction Ff = µFn
\Ff = 0.75*455.65
\Ff = 341.74 N
\Resultant force(F) is remained shoved force when frictional loss forces subtracted in horizontal direction.
\F = Fsh-Ff
\a = acceleration of Book box.
\ma = 352.24 - 341.74
\21.33a = 10.5
\a = 0.492
\From motion equations
\s = ut + ½at²
\Box starts from rest.So initial velocity u = 0
\8 = 0 + ½(0.492)t²
\t² = 8*2/0.492
\t = 32.52.sec
\Solution is 32.52 sec
\