Question

A spring (k = 802 N/m) is hanging from the ceiling of an elevator, and a 5.0-kg object is attached to the lower end. By how much does the spring stretch (relative to its unstrained length) when the elevator is accelerating upward at a = 0.41 m/s2?

Answer 1

Answer:

0.00256 m

Explanation:

For this case, we should use the Hook Law for a spring:

F= -Kx , the negative sign indicates that the force that tries to “restore” the original status of the spring, and is against the force that causes the spring´s displacement

This is, the force F required to stretch an elastic object (for example, a metal spring), is directly proportional the extension “x” of the spring

“x” can be the extension or compression of the spring, and “K” is a constant (in units of N/m)

We also know that, according to Newton´s Law:

F=m*a

m= mass

a= acceleration

Then:

m*a= -Kx

Finally, the spring has an original length of L₀, so:

- If the spring is compressed, the final strength will be: L = L₀ – x

- If the spring is extended, the final strength will be: L = L₀ + x

According to the statement:

K = 802 N/m

a = 0.41 m/s2

m = 5 Kg

Then:  

x =- (m*a)/K = (5x0.41)/802 = 0.00256 m (we do not consider the negative sign, due to above explanation: it only indicates that the restoration force is always against the force imposed on the spring )

So, if the spring has an original length of L₀, when the elevator is accelerating upwards, the spring will stretch from L₀ to (L₀ – 0.00256) m