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