The Work–Energy Theorem

The work–energy theorem is the bridge between forces and motion: the net work done on an object equals the change in its kinetic energy:

$W_{\text{net}} = \Delta\text{KE} = \text{KE}_f - \text{KE}_i$

It's a direct consequence of Newton's Second Law and one of the most useful tools in mechanics, because it lets you skip directly from a force/distance setup to a final speed without ever computing the time taken.

The Formula for Constant Force

For a constant force $F$ acting over a parallel displacement $d$:

$W = F\,d$

Combining with the theorem:

$F\,d = \tfrac{1}{2}mv_f^{2} - \tfrac{1}{2}mv_i^{2}$

If the object starts from rest ($v_i = 0$), this simplifies to:

$F\,d = \tfrac{1}{2}mv_f^{2}$

$v_f = \sqrt{\dfrac{2 F\,d}{m}}$

Step-by-Step Solution

Given: $F = 10\;\text{N}$, $m = 2\;\text{kg}$, $d = 5\;\text{m}$, $v_i = 0$.

Find: Final speed $v_f$.

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Step 1 — Compute the work done.

$W = F\,d = 10 \times 5 = 50\;\text{J}$

Step 2 — Apply the work–energy theorem.

$W = \tfrac{1}{2}mv_f^{2} - \tfrac{1}{2}mv_i^{2} = \tfrac{1}{2}mv_f^{2} - 0$

$50 = \tfrac{1}{2}(2)\,v_f^{2}$

$50 = v_f^{2}$

$v_f = \sqrt{50} \approx 7.071\;\text{m/s}$

Step 3 — Cross-check using kinematics.

The acceleration is $a = F/m = 10/2 = 5\;\text{m/s}^{2}$. Using $v^{2} = v_0^{2} + 2 a d$:

$v_f^{2} = 0 + 2(5)(5) = 50$

$v_f = \sqrt{50} \approx 7.071\;\text{m/s} \;\;\checkmark$

Both methods agree. The work–energy theorem is faster because it skips the time variable entirely.

Step 4 — How long does it take?

If you needed the time, use $v_f = at$:

$t = \dfrac{v_f}{a} = \dfrac{7.071}{5} \approx 1.414\;\text{s}$

Step 5 — Sanity check the final KE.

$\text{KE}_f = \tfrac{1}{2}(2)(50) = 50\;\text{J} = W \;\;\checkmark$

All 50 J of work converted into kinetic energy. None lost (no friction).

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Answer:

$\boxed{\;v_f = \sqrt{\dfrac{2Fd}{m}} = \sqrt{50} \approx 7.071\;\text{m/s}\;}$

The work done by the 10 N force over 5 m equals 50 J, which becomes the box's kinetic energy at the end.

Try It

• Adjust F, m, and d sliders.
• Notice that the final speed scales as $\sqrt{Fd/m}$ — quadrupling the distance only doubles the speed.
• Watch the work bar grow as the box moves; it equals the area under the constant-force curve.
• Compare with the kinematics approach — they always agree.