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

$U = \frac{1}{2}kx^2$

now it is given to us that

$U = 3500 J$

$x = 1.5 m$

now we will plug in data in above equation

$3500 = \frac{1}{2} k(1.5)^2$

$k = \frac{2 \times 3500}{1.5^2}$

now we will have

$k = 3111 N/m$