To solve this question using the work-energy theorem, let's outline what the theorem states and how it applies to the formulas provided:
The work-energy theorem states that the work done by the net force acting on an object is equal to the change in the kinetic energy of that object. The formula for kinetic energy [tex]\( KE \)[/tex] is given by:
[tex]\[ KE = \frac{1}{2}mv^2 \][/tex]
where [tex]\( m \)[/tex] is the mass of the object and [tex]\( v \)[/tex] is its velocity.
To find the work [tex]\( W \)[/tex] done, we need to calculate the change in kinetic energy [tex]\( \Delta KE \)[/tex]. If an object changes its velocity from [tex]\( v_i \)[/tex] (initial velocity) to [tex]\( v_f \)[/tex] (final velocity), the change in kinetic energy can be computed as follows:
[tex]\[ \Delta KE = KE_{final} - KE_{initial} \][/tex]
Substituting the expressions for the final and initial kinetic energies, we get:
[tex]\[ \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 \][/tex]
This can be rewritten to:
[tex]\[ \Delta KE = \frac{1}{2}m(v_f^2 - v_i^2) \][/tex]
Therefore, the work done [tex]\( W \)[/tex], which equals the change in kinetic energy [tex]\( \Delta KE \)[/tex], is:
[tex]\[ W = \frac{1}{2}m(v_f^2 - v_i^2) \][/tex]
Now, let's match this result with the provided options:
1. [tex]\( W = \Delta KE = 2m(v_i^2 - v_f^2) \)[/tex]
2. [tex]\( W = \Delta KE = \frac{1}{2}m(v_i^2 - v_f^2) \)[/tex]
3. [tex]\( W = \Delta KE = 2m(v_f^2 - v_i^2) \)[/tex]
4. [tex]\( W = \Delta KE = \frac{1}{2}m(v_f^2 - v_i^2) \)[/tex]
From inspecting these expressions, option 4:
[tex]\[ W = \Delta KE = \frac{1}{2}m(v_f^2 - v_i^2) \][/tex]
matches the derived formula exactly. Thus, the correct formula according to the work-energy theorem for determining the amount of work done is:
[tex]\[ W = \Delta KE = \frac{1}{2}m(v_f^2 - v_i^2) \][/tex]
Hence, the answer to the question is the fourth option.
