A spring of unstretched length l has a mass m with one end fixed to a rigid support. Assuming spring to be made of a uniform wire, the kinetic energy possessed by if its free end is pulled with uniform velocity v is a. $\frac{1}{2}m{v}^2$ b. $\frac{1}{3}m{v}^2$ c. $m{v}^2$ d. $\frac{1}{6}m{v}^2$

Certainly! Here is an example of a step-by-step solution for a math problem: **Problem:** Simplify the expression $\frac{(2x^2-4x+6)}{(x-3)}$ **Solution:** Step 1: Factor the numerator. $2x^2-4x+6=2(x^2-2x+3)$ Step 2: Rewrite the fraction with the factored numerator. $\frac{2(x^2-2x+3)}{(x-3)}$ Step 3: Factor the quadratic expression in the numerator. $x^2-2x+3=(x-1)(x-2)$ Step 4: Substitute the factored quadratic into the numerator of the fraction. $\frac{2(x-1)(x-2)}{(x-3)}$ Step 5: Cancel any common factors between the numerator and denominator. $\frac{2(x-1)(x-2)}{(x-3)}=\frac{2(x-1)}{1}\cdot \frac{(x-2)}{(x-3)}=\boxed{2\frac{(x-1)}{(x-3)}(x-2)}$ And that's it! The step-by-step solution breaks down the problem into manageable steps, making it easier for students to follow and understand.