understand the concept of work, and recall and use work done = force × displacement in the direction of the force

Energy Conservation

Learning Objective

Students will understand the concept of work and be able to recall and use the formula

\$W = \vec F \cdot \vec s = Fs\cos\theta\$

where W is the work done by a constant force F acting through a displacement s at an angle θ to the direction of the force.

1. What is Work?

Work is the transfer of energy that occurs when a force causes a displacement. It is a scalar quantity and can be positive, negative or zero depending on the relative direction of the force and the displacement.

2. Deriving the Work‑Force‑Displacement Relationship

1. Consider a constant force F acting on an object.
2. The displacement vector is s. The angle between the force and displacement is θ.
3. The component of the force in the direction of the displacement is F\cosθ.
4. Multiplying this component by the magnitude of the displacement gives the work:

   \$W = (F\cos\theta)s = Fs\cos\theta\$

5. In vector notation this is the dot product: \$\vec F\cdot\vec s\$

3. Sign of Work

• Positive work: 0° ≤ θ < 90° – force has a component in the direction of motion (e.g., pushing a sled forward).
• Negative work: 90° < θ ≤ 180° – force opposes the motion (e.g., friction, braking).
• Zero work: θ = 90° – force is perpendicular to displacement (e.g., centripetal force in uniform circular motion).

4. Units

5. Example Problems

Example 1 – Horizontal Push

A student pushes a 5 kg crate across a frictionless floor with a constant horizontal force of 20 N over a distance of 3 m. Calculate the work done on the crate.

1. Identify the quantities: \(F = 20\;\text{N}\), \(s = 3\;\text{m}\), \(\theta = 0^\circ\) (force parallel to displacement).
2. Apply the formula:

   \$W = Fs\cos\theta = (20\;\text{N})(3\;\text{m})\cos0^\circ = 60\;\text{J}\$

3. Result: The work done is 60 J (positive, because the force aids the motion).

Example 2 – Inclined Plane

A block of mass 2 kg is pulled up a smooth incline that makes an angle of \(30^\circ\) with the horizontal. The pulling force is 15 N directed parallel to the incline, and the block moves 4 m along the incline. Find the work done by the pulling force.

1. Force and displacement are parallel to the incline, so \(\theta = 0^\circ\).
2. Work:

   \$W = Fs\cos0^\circ = (15\;\text{N})(4\;\text{m}) = 60\;\text{J}\$

3. The work done by gravity is negative:

   \$W_g = -m g h = - (2\;\text{kg})(9.81\;\text{m s}^{-2})(4\sin30^\circ) \approx -39.2\;\text{J}\$

4. Net work on the block is \(60\;\text{J} - 39.2\;\text{J} = 20.8\;\text{J}\), which equals the increase in kinetic energy.

6. Using Work to Apply the Conservation of Energy

When only conservative forces (e.g., gravity, elastic spring) do work, the total mechanical energy is conserved:

\$Ki + Ui + W{\text{nc}} = Kf + U_f\$

where \(W{\text{nc}}\) is the work done by non‑conservative forces (e.g., friction). If \(W{\text{nc}} = 0\), then \(Ki + Ui = Kf + Uf\).

7. Common Misconceptions

• “Work is done whenever a force is applied.” – Work requires *displacement* in the direction of the force.
• “If an object moves, work must have been done on it.” – The net work could be zero if positive and negative works cancel (e.g., constant speed on a level surface with friction).
• “Work is a vector.” – Work is a scalar; its sign indicates direction relative to displacement, not a vector direction.

8. Summary Table

9. Practice Questions

1. Calculate the work done by a 10 N force acting at 60° to the horizontal when an object moves 5 m horizontally.
2. A 0.5 kg ball is thrown vertically upward with an initial speed of 8 m s⁻¹. Determine the work done by gravity when the ball reaches its highest point.
3. Explain why the work done by the normal reaction force on a block sliding down a frictionless incline is zero.

10. Key Take‑aways

• Work quantifies energy transfer: \(W = \vec F\cdot\vec s = Fs\cos\theta\).
• Only the component of the force parallel to the displacement contributes to work.
• Positive work adds energy to the system; negative work removes energy.
• When only conservative forces do work, mechanical energy is conserved.