A horizontal spring with spring constant 200N/m is compressed by 15cm and used to launch a 2kg box across a frictionless horizontal surface. After the box travels some distance, the surface becomes rough. The coefficient of kinetic friction of the box on the surface is 0.2. Use work and energy to find how far the box slides across the rough surface before stopping.

Explanation:

Given that,

Spring constant k=200N/m

Compression x = 15cm = 0.15m

Attached mass m =2kg

Coefficient of kinetic friction uk= 0.2

The energy in the spring is given as

U =½kx²

U = ½ × 200 × 0.15²

U = 2.25J

Force in the spring is given by Hooke's law

F = ke

F = 200×0.15

F = 30N

The weight of body which is equal to the normal is give as

W = mg

W = 2 × 9.81

W = 19.62N

W = N = 19.62 Newton's 2nd Law

From law of friction,

Fr = uk•N

Fr = 0.2 × 19.62

Fr = 3.924

Using newton second law again

Fnet = F - Fr

Fnet = 30 - 3.924

Fnet = 26.076

Work done by net force is given as

W = Fnet × d

W = 26.076d

Then, the work done by this net force is equal to the energy in the spring

W = U

26.076d = 2.25

d = 2.25/26.076

d = 0.0863m

Which is 8.63cm

So the box will slide 8.63cm before stopping

  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