Physics Spring Question

Question

In order to determine the acceleration of an elevator, you hang a 0.57-kg mass from the end of a spring with spring constant k = 0.93 N/cm. When the elevator starts to accelerate, you measure the length of the spring and note that it has stretched 3.82 cm from its initial unstretched length (when the mass was not hanging from it). What is the magnitude of the elevator's acceleration?

Answer 1

Givens:

m=.57kg

k=0.93N/cm

Δx=-3.82cm (because it stretches down)

g=-9.81m/s²

Equations:

F_s=-kΔx

F_g=mg

F_net=ma

Draw a free body diagram for once the elevator is accelerating and create force equation for what forces are actually acting on the mass itself

F_net=F_s+F_g

ma=-kx+mg

a=(-kx+mg)/m

a=((-0.93N/cm)(-3.82cm)+(.57kg)(-9.81m/s²))/.57=-3.58m/s²

Since it only asks for magnitude the answer is 3.58m/s² but we know the acceleration is down. It is still important to make sure we use the correct signs in the equation because gravity and the spring force pull in opposite directions and if we didn't use the correct signs we could get too large an acceleration.

If we are skeptical that the elevator is moving down we could use the conditions before the elevator is moving to find the original stretch of the spring and then compare it and we can see that it is closer to its equilibrium position after the elevator starts moving meaning that the Force of the spring is less than the force of gravity once the elevator starts moving. If the force of the spring were greater it would be stretched further from equilibrium and provide an upward net force (upward acceleration)