Question

When a man increases his speed by 2 m/s, he finds that his kinetic energy is doubled, the & original speed of the man is:

• A. $2(\sqrt{2}-1)$ m/s
• B. $2(\sqrt{2}+1)$ m/s
• C. 4.5 m/s
• D. none of these

Answer 1

Answer: (B)

$KE=K=\frac{1}{2}m{v}^2$ Given: $v_2=v_1+2$ $\therefore$ $\frac{K_1}{K_2}={\left(\frac{v_1}{v_2}\right)}^2$ Or $\frac{K_1}{K_2}={\left(\frac{v_1}{v_2}\right)}^2$ Or $\frac{K_1}{2K_2}=\frac{v_1^2}{{(v_1+2)}^2}$ Or $v_1^2+4v_1+4=2v_1^2$ Or $v_1^2-4v_1-4=0$ $v_1=\frac{4\pm \sqrt{16+6}}{2}$ $=\frac{4\pm \sqrt{32}}{2}$ $=2(\sqrt{2}+1)m/s$