5.3. WORK

In physics, work has a specific definition. Can anyone tell me what that is?

Isn't work just when we do something and get tired?

That's a common misconception! Physically, work is defined as the product of force and displacement in the direction of that force. We express this as W = F * d * cos(θ).

So, if there's no movement, there is no work done?

Exactly! If displacement is zero, work is zero, no matter how much force was applied. Remember, if you're pushing against a wall and it doesn't move, you're not doing any work. Work relates directly to displacement!

What about when the direction of force is different from the direction of displacement?

Good question! If the angle θ is between 0° and 90°, work is positive. When it’s between 90° and 180°, the work done is negative! This means work done against the movement of an object.

So how do we know if work is positive or negative?

Remember this simple rule: positive work helps move the object in the direction of the force, while negative work acts to slow it down. Let’s summarize: Work = Force × Displacement × cos(θ).

Now let’s connect what we've learned about work to kinetic energy. Who remembers what kinetic energy is?

Isn't that the energy an object has due to its motion?

Exactly! The kinetic energy of an object is given by K = 1/2 mv². The work-energy theorem states that the work done by a net force on an object equals the change in its kinetic energy: K_f - K_i = W.

So, every time we do work on an object, we change its kinetic energy?

Correct! For example, if you push a car and it starts moving, the work you apply increases its kinetic energy. Does anyone recognize situations where work results in no change in energy?

If an object is moving at a constant speed, does that mean no net work is done?

Yes! If the forces acting on it are balanced, no net work is done. Let’s summarize: Work changes kinetic energy!

Let’s explore some examples together. Suppose a force of 10 N is applied over a distance of 5 m in the same direction. How much work is done?

I think it’s W = 10 N × 5 m, so W = 50 J.

Great! Now, if the same force is applied at a 60° angle to the direction of motion, how would we calculate it?

We would use W = F * d * cos(θ). So W = 10 N * 5 m * cos(60°), which is 25 J!

Exactly! Remember the cosine function lets us find the effective component of force acting in the direction of displacement. Let’s look at another real-world example: a weight lifter holds a weight steadily. Is any work done?

No, because the weight hasn't moved!

Perfect! This emphasizes our understanding that work requires displacement. To summarize, work done depends on force, displacement and angle!

Now we will dive deeper into calculations involving work. Let’s say we have an object being pushed along a rough surface. If a force of 20 N is applied at an angle of 30° while moving 4 m, how would we calculate the work done?

We would calculate W = F * d * cos(θ), which would be W = 20 N * 4 m * cos(30°).

You got it! Cos(30°) gives approximately 0.866, so the work done is around 69.28 J. What happens if the object experiences friction?

Then we would need to subtract the work done by friction from the total work calculated!

Exactly! Always make sure to consider all forces acting on the object. We have learned that the total work considers contributions from all forces, including friction!

Can we do another example for practice?

Of course! How about everyone calculates work done when moving an object vertically against gravity?