Sure, let's solve the problem step-by-step.
First, recall the formula for kinetic energy:
\(\[ KE = \frac{1}{2}mv^2 \]\)
where:
- \(\( KE \)\) is the kinetic energy,
- \(\( m \)\) is the mass of the body,
- \(\( v \)\) is the speed of the body.
Given that the kinetic energy of the body is \(\( E \)\) when the speed is \(\( v \)\):
\(\[ E = \frac{1}{2}mv^2 \]\)
We want to find the new speed (\(\( v_{\text{new}} \)\)) of the body when its kinetic energy is \(\( 4.0E \)\).
Let's set up the situation for the new kinetic energy:
\(\[ 4.0E = \frac{1}{2}mv_{\text{new}}^2 \]\)
We can substitute \(\( E = \frac{1}{2}mv^2 \)\) into the equation:
\(\[ 4.0 \times \frac{1}{2}mv^2 = \frac{1}{2}mv_{\text{new}}^2 \]\)
This simplifies to:
\(\[ 2mv^2 = \frac{1}{2}mv_{\text{new}}^2 \]\)
To solve for \(\( v_{\text{new}} \)\), we first isolate \(\( v_{\text{new}}^2 \)\):
\(\[ 2mv^2 = \frac{1}{2}mv_{\text{new}}^2 \]\)
\(\[ 2v^2 = \frac{1}{2}v_{\text{new}}^2 \]\)
\(\[ 4v^2 = v_{\text{new}}^2 \]\)
Take the square root of both sides to solve for \(\( v_{\text{new}} \)\):
\(\[ v_{\text{new}} = \sqrt{4v^2} \]\)
\(\[ v_{\text{new}} = 2v \]\)
So, the speed of the body when its kinetic energy is \(\( 4.0E \)\) is:
\(\[ v_{\text{new}} = 2v \]\)
The correct answer is \(\( \boxed{2.0v} \)\).
