Question

A cyclist doubles his speed. How will his kinetic energy change: increase, decrease or remain same?

Answer 1

As Kinetic energy K.E. = 1/2 Mv₂
Since speed is doubled, its square will become 4 times
∴ K.E. increases i.e. become 4 times.
Or
K.E j =  Mv₂
When speed is doubled New speed v₁ = (2 v)
New K.${\displaystyle E_2}$= ${\displaystyle \frac{1}{2}}$M${\displaystyle V_1^2}$

∴ New K.${\displaystyle E_2}$= ${\displaystyle \frac{1}{2}}$ M (2v${\displaystyle {)}^2}$

             K.${\displaystyle E_2}$=${\displaystyle \frac{1}{2}}$ M 4${\displaystyle V^2}$=4${\displaystyle \left[\frac{1}{2}M{V}^2\right]}$

     (¡¡) ÷ (¡)

            =${\displaystyle \frac{K{E}_2}{K{E}_1}}$ =${\displaystyle \frac{4\left[\frac{1}{2}M{V}^2\right]}{\left[\frac{1}{2}M{V}^2\right]}}$=${\displaystyle \frac{4}{1}}$

            =${\displaystyle \frac{K{E}_2}{K{E}_1}}$

        New K.${\displaystyle E{.}_2}$=4 Times K.${\displaystyle E_1}$