If the coefficient of kinetic friction between a 29 kg crate and the floor is 0.38, what horizontal force is required to move the crate at a steady speed across the floor?

Let F be the magnitude of the force needed to keep the box sliding at constant speed, and f the magnitude of kinetic friction. Then by Newton's second law, the net horizontal force on the crate is

F - f = 0

so that

F = f

f is proportional to the magnitude of the normal force n,

f = 0.38n

The net vertical force is

n - mg = 0

which tells us that

n = mg = (29 kg) (9.8 m/s²) ≈ 284 N

Then the required force must have magnitude

F = f = 0.38 (284 N) ≈ 110 N

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂