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