Questions: An engineer is designing a spring to be placed at the bottom of an elevator shaft. If the elevator cable should break when the elevator is at a height h above the top of the spring, calculate the value that the spring stiffness constant (k) should have so that passengers undergo an Let M be the total mass of the elevator and passengers. mg(h+x)=1 / 2 k x^2 F=k x=m cdot a k x=m cdot g mg(h+(mg/k))=1 / 2 k(mg/k)^2

Solution Steps

• Use the given energy conservation equation: \[ mg(h+x) = \frac{1}{2} k x^2 \]
• Here, \( h \) is the height above the spring, \( x \) is the compression of the spring, \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( k \) is the spring constant.

• Use the force equation given: \[ F = k x = m \cdot a \]
• Given that the acceleration \( a \) is \( 6g \): \[ k x = m \cdot 6 \cdot g \]

• Solve the force equation for \( x \): \[ x = \frac{m \cdot 6 \cdot g}{k} \]

• Substitute \( x = \frac{6mg}{k} \) into the energy conservation equation: \[ mg \left( h + \frac{6mg}{k} \right) = \frac{1}{2} k \left( \frac{6mg}{k} \right)^2 \]

• Simplify the left-hand side: \[ mg \left( h + \frac{6mg}{k} \right) \]
• Simplify the right-hand side: \[ \frac{1}{2} k \left( \frac{6mg}{k} \right)^2 = \frac{1}{2} k \cdot \frac{36m^2g^2}{k^2} = \frac{18m^2g^2}{k} \]

• Equate the simplified expressions: \[ mg \left( h + \frac{6mg}{k} \right) = \frac{18m^2g^2}{k} \]
• Distribute \( mg \): \[ mgh + \frac{6m^2g^2}{k} = \frac{18m^2g^2}{k} \]
• Isolate \( k \): \[ mgh = \frac{18m^2g^2}{k} - \frac{6m^2g^2}{k} \] \[ mgh = \frac{12m^2g^2}{k} \] \[ k = \frac{12m^2g^2}{mgh} \] \[ k = \frac{12mg}{h} \]

Final Answer