A spring with mas 1.0 kg has a natural length of 0.53 m. A force of 4.25 N is required to stretch it to a length of 0.70 m. If the spring is stretched to a length of 0.66 m then released with a downward speed of 2.1 m/s. Find the maximum displacement of the mass.
You mean the spring's mass is 1 kg right? Or do you mean that a 1 kg mass is attached to it?
the spring is 1 kg
Okay.. Hmm. Seems like we will need integration. And what does 'downward' mean? Where and how is the spring placed initially? Is there a diagram alongwith the question?
pull the spring down
do you have the answer?
Ennergy conservation -> apply
I'm sure you're familiar with formula such as F=kx, Elastic Potential Energy=(1/2)kx^2... So first start with F=kx. To find k. given F=4.25 when spring is extended to 0.70m. so x=0.70-0.53. Then use the principle of conservation of energy. When the mass is released, it contains Elastic Potential Energy and Kinetic Energy. E(before)=(1/2)kx^2 + (1/2)mv^2. we found k, x=0.66-0.53,m=1, v=2.1. After releasing and reaching the maximum displacement, means the velocity of the mass=0 and the spring is extended to the maximum. E(after)=(1/2)ks^2 where s is the maximum displacement. Equate E(before) and E(after) to find s.
I have to use differential equations
Sorry I don't know the differential equation method, I just know the physics method.
okay thanks for trying tho
Join our real-time social learning platform and learn together with your friends!