A box of books weighing 209 N is shoved across the floor

Question

by a force of 430 N exerted downward at an angle of 35° below the horizontal.?

a. If µk between the box and the floor is 0.57, how long does it take to move the box 5 m, starting from rest? (If the box will not move, enter 0.)

b. If µk between the box and the floor is 0.75, how long does it take to move the box 8 m, starting from rest? (If the box will not move, enter 0.)

 

 

 

Answer 1

(a)

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 = 0.57

Book box moved distance s = 5 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 = 0.57*455.65

Ff = 259.72 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 - 259.72

21.33a = 92.52

a = 4.337

From motion equations

s = ut + ½at²

Box starts from rest.So initial velocity u = 0

5 = 0 + ½(4.337)t²

t² = 5*2/4.337

t = 1.52 sec

Solution is 1.52 sec

Answer 2

(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