Question

A stockroom worker pushes a box with mass 11.5kg on a horizontal surface with a constant speed of 3.60m/s . The coefIficient of kinetic friction between the box and the surface is 0.220.

Part A

What horizontal force must be applied by the worker to maintain the motion?

Part B
If the force calculated in part A is removed, how far does the box slide before coming to rest?

Answer 1

Answer:a) 24.794N

b) 3.67m

Step-by-step explanation:

F>f

fk= kinetic coefficient × n

EFy = may =0

n - mg = 0

n = mg==> 11.5 × 9.8 = 112.7N

fk= kinetic coefficient × n= 0.220 × 112.7

fk= 24.794N

b) fk= 0, ◇x= ?

◇x = Vt + 1/2(at)^2

But V= 0

t= mv/fk = 11.5 × 3.60/24.794

t= 41.4/24.794 = 1.67 seconds

Efx= max

fx/m =ax = 24.794/11.5 = 2.156m

◇x= 0(1.67) + 1/2 (2.156)(1.69)^2

◇x = 1.67 + 1/2(6.013)

◇x = 1.67 + 2 = 3.67m

Answer 2

Part A = What horizontal force must be applied by the worker to maintain the motion = 24.82N

Part B = how far does the box slide before coming to rest = 3.00m

Step-by-step explanation:

The detailed calculation is shown in the attached file.