(III) An engineer is designing a spring to be placed at the bottom of an elevator shaft. If the elevator cable breaks when the elevator is at a height h above the top of the spring, calculate the value that the spring constant k should have so that passengers undergo an acceleration of no more than 5.0 g when brought to rest. Let M be the total mass of the elevator and passengers.

12Mg/h

Explanation:

Let the spring is compressed by a distance x,before the lift stops,then

Mg(h+x)= 1/2 kx^2 1

Kx - Mg = M ( 5g ) 2

Make x the subject in equation 2

Kx = 5Mg + Mg

Kx = 6Mg

x = 6Mg/k 3

Put equation 3 into 1

Mg ( h + x ) = 1/2 kx^2

Mgh + Mgx = 1/2kx^2

Mgh + Mg × 6Mg/k = 1/2k × ( 6Mg/k )^2

Mgh + Mg× 6Mg/k = 1/2k 36M^2g^2/ k^2

h =18Mg/k - 6Mg/h

K = 12Mg/h

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz

grav potential energy