An engineer is designing a spring to be placed at the bottom of an elevator shaft. if the elevator cable should happen to break when the elevator is at a height h = 12.5 m above the top of the spring, calculate the value the spring constant k should have so that passengers undergo a maximum acceleration of 5.3g. take the mass of the elevators plus passengers to be m = 1439 kg.

Explanation:

As per energy conservation we know that

initial total gravitational potential energy = final spring potential energy

so we have

also we know that maximum acceleration will be 5.3 g

so it is given as

so we have

  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

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