Recall and use the equation for mechanical working W = F d = ΔE

1.7.2 Work

Learning Objective

Recall and use the equation for mechanical working:

\$W = F d = \Delta E\$

What is Work?

Work is done when a force causes an object to move through a distance in the direction of the force. It is a scalar quantity that transfers energy to or from a system.

Mathematical Formulation

• General definition: \$W = \vec F \cdot \vec d = F d \cos\theta\$
• When the force is parallel to the displacement (\$\theta = 0^\circ\$): \$W = F d\$
• When the force is opposite to the displacement (\$\theta = 180^\circ\$): \$W = -F d\$ (negative work)

Units and Symbols

Key Points to Remember

1. Work is only done if there is a component of force in the direction of motion.
2. Zero displacement or a force perpendicular to the motion results in zero work.
3. Positive work adds energy to the system; negative work removes energy.
4. The magnitude of work equals the change in kinetic energy (Work‑Energy Theorem): \$W = \Delta KE\$.

Common Situations

• Pulling a sled horizontally: \$W = F d\$ (positive work if pulling forward).
• Lifting a box vertically: \$W = m g h\$ (positive work against gravity).
• Holding a weight stationary: \$d = 0\$, so \$W = 0\$ even though a force is applied.
• Moving a block on a frictionless surface with a force at 30° to the horizontal: \$W = F d \cos30^\circ\$.

Suggested Diagram

Worked Example

Problem: A student pushes a 5.0 kg crate with a constant horizontal force of 20 N over a distance of 3.0 m. Calculate the work done on the crate.

Solution:

1. Identify the force component in the direction of motion: \$F = 20\ \text{N}\$ (already horizontal).
2. Displacement: \$d = 3.0\ \text{m}\$.
3. Apply the formula: \$W = F d = 20\ \text{N} \times 3.0\ \text{m} = 60\ \text{J}\$.
4. Since the force and displacement are in the same direction, the work is positive: \$W = +60\ \text{J}\$.

Practice Questions

1. A 10 kg block is lifted vertically 2.5 m. Calculate the work done against gravity. (Take \$g = 9.8\ \text{m s}^{-2}\$.)
2. A force of 15 N acts at \$45^\circ\$ to the horizontal on a cart that moves 4.0 m horizontally. Find the work done by the force.
3. A person holds a 30 N weight stationary for 10 s. How much work is done on the weight?
4. A car accelerates from rest to 20 m s\$^{-1}\$ in 5 s. Its mass is 800 kg. Assuming the net force is constant, calculate the work done on the car.

Summary Table

Key Take‑away

Work quantifies how much energy is transferred by a force acting through a distance. Remember the simple form \$W = F d\$ when the force and motion are aligned, and always check the direction of the force relative to the displacement to determine the sign of the work.