Does the theorem work if the net work is zero?

Yes! If the net work is zero, then $\Delta K = 0$. This means the kinetic energy doesn't change, so the object maintains a constant speed. This happens when the pushing force perfectly balances the friction force.

What does negative work mean here?

Negative work means the net force is acting in the opposite direction of the object's motion. According to the theorem, negative net work results in a negative change in kinetic energy, meaning the object slows down.

Does this theorem apply to potential energy too?

The basic work-energy theorem ($W_{\text{net}} = \Delta K$) specifically connects net work to kinetic energy. While work done by conservative forces (like gravity) can be rewritten as a change in potential energy, the fundamental theorem itself is strictly about kinetic energy.

What is the work-energy theorem?

The work-energy theorem is a fundamental physics principle stating that the net work done on an object is exactly equal to the change in its kinetic energy. In equation form, it is written as $W_{\text{net}} = \Delta K$ . Think of your bank account: the net money you deposit or withdraw (work) exactly equals the change in your account balance (kinetic energy). If you push a shopping cart forward (doing positive work), its speed increases. If friction drags against the cart (doing negative work), its speed decreases.

What it is: Defining the terms

To use this theorem, we need to understand its two pieces. Net work ( $W_{\text{net}}$ ) is the sum of all the work done by every force acting on the object. Kinetic energy ( $K$ ) is the energy an object has due to its motion, calculated using the formula $K = \frac{1}{2}mv^2$ . The theorem simply says that whenever forces act on an object over a distance, the total effort (work) translates directly into a change in the object's motion energy.

Why it works: The connection to forces

This theorem isn't just a coincidence; it is born directly from Newton's second law ( $F = ma$ ) and the equations of motion (kinematics). When a net force pushes an object over a certain distance, it causes the object to accelerate. That acceleration means the object's velocity is changing. By multiplying the net force by the distance, we are effectively calculating exactly how much the kinetic energy has changed.

Where students slip: Net work vs. single work

The most common mistake students make is calculating the work done by just one force (like a person pushing a box) and setting that equal to the change in kinetic energy. You must account for all forces! If you push a box but friction is pushing back, you must subtract the work done by friction to find the true net work. Only the net work equals $\Delta K$ .

Worked through

A 1000 kg car is traveling at 10 m/s. The driver applies the brakes, which exert a steady net frictional force of -2000 N over a distance of 20 m. What is the final speed of the car?

First, we calculate the net work done on the car. $W_{\text{net}} = F_{\text{net}} \times d = -2000 \text{ N} \times 20 \text{ m} = -40,000 \text{ J}$ . Next, we find the initial kinetic energy: $K_i = \frac{1}{2}mv^2 = \frac{1}{2}(1000 \text{ kg})(10 \text{ m/s})^2 = 50,000 \text{ J}$ . Using the work-energy theorem, $W_{\text{net}} = K_f - K_i$ . Plugging in our numbers: $-40,000 \text{ J} = K_f - 50,000 \text{ J}$ . Solving for final kinetic energy gives $K_f = 10,000 \text{ J}$ . Finally, we use the kinetic energy formula to find the final speed: $10,000 = \frac{1}{2}(1000)v_f^2$ . This simplifies to $10,000 = 500v_f^2$ , so $v_f^2 = 20$ . Taking the square root, $v_f \approx 4.47 \text{ m/s}$ .

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Where this comes from: OpenStax University Physics Volume 1 · Khan Academy AP Physics 1