Question

A spring stores 3500 j of energy when it is stretched 1.5 meters. What is the spring constant of this spring? A) 4667 N/m B) 778 N/m C) 3938 N/m D) 3111 N/m

Answer 1

As we know that spring potential energy is given by the formula

`U = (1)/(2)kx^2`

now it is given to us that

`U = 3500 J`

`x = 1.5 m`

now we will plug in data in above equation

`3500 = (1)/(2) k(1.5)^2`

`k = (2 * 3500)/(1.5^2)`

now we will have

`k = 3111 N/m`