Work IB DP Physics Study Notes - 2025 Syllabus
Work IB DP Physics Study Notes
Work IB DP Physics Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on IB Physics syllabus with Students should understand
- that work done by a force is equivalent to a transfer of energy
- that energy transfers can be represented on a Sankey diagram
- that work W done on a body by a constant force depends on the component of the force along the line of displacement as given by W = Fs cos θ
- that work done by the resultant force on a system is equal to the change in the energy of the system
Standard level and higher level: 8 hours
Additional higher level: There is no additional higher level content
- IB DP Physics 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Physics 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Physics 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Physics 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
Work Done by a Force is Equivalent to a Transfer of Energy
In physics, work is the mechanism through which energy is transferred into or out of a system by a force. When a force acts upon an object and displaces it, energy is transferred from the agent applying the force to the object.
The amount of energy transferred depends on both the magnitude of the force and the distance over which it acts. The object either gains or loses energy depending on the direction of the force relative to its motion.
This transfer of energy may result in:
- An increase in kinetic energy (object speeds up)
- An increase in potential energy (object raised against gravity)
- Thermal energy (e.g., due to friction)
Example:
A person pushes a crate with a force of \( 50 \, \text{N} \) over a distance of \( 3 \, \text{m} \) on a frictionless surface.
Calculate the work done and explain how energy is transferred.
▶️ Answer/Explanation
Since the surface is frictionless and the force is applied in the direction of motion:
\( W = F \cdot s = 50 \cdot 3 = \boxed{150 \, \text{J}} \)
The object gains 150 J of kinetic energy, transferred from the person’s muscles (chemical energy) to the crate. This is a direct example of work as a transfer of energy.
Energy Transfers Can Be Represented on a Sankey Diagram
A Sankey diagram is a flow chart used to visually represent the energy transfers within a system. The arrows in a Sankey diagram have widths proportional to the amount of energy they represent.
Key features:
- Input energy: The total energy supplied to the system.
- Useful output energy: Energy used for the intended purpose (e.g., kinetic energy in a motor).
- Wasted energy: Energy lost to the surroundings, usually as heat or sound.
Sankey diagrams obey the law of conservation of energy:
Total input energy = useful energy output + wasted energy
Example:
An electric motor receives 100 J of electrical energy. It outputs 70 J as mechanical energy, and the rest is lost as heat.
▶️ Answer/Explanation
Input Energy = \( 100 \, \text{J} \)
Useful Mechanical Output = \( 70 \, \text{J} \)
Wasted Heat Energy = \( 100 – 70 = 30 \, \text{J} \)
This can be represented with a Sankey diagram as:
→ [100 J] → [70 J] (useful) ↘ [30 J] (wasted as heat)
Efficiency = \( \frac{70}{100} \times 100\% = \boxed{70\%} \)
Work \( W \) Done on a Body by a Constant Force
When a constant force acts on an object and causes it to move, the amount of work done is calculated by:
\( W = F s \cos \theta \)
Where:
- \( F \): Magnitude of the applied force (in newtons)
- \( s \): Displacement of the object (in meters)
- \( \theta \): Angle between the force direction and displacement
The cosine factor ensures that only the component of the force along the displacement direction contributes to the work done.
Special Angle Cases:
- \( \theta = 0^\circ \): Force in direction of displacement → \( \cos 0^\circ = 1 \) → \( W = Fs \) (maximum work)
- \( \theta = 90^\circ \): Force perpendicular to displacement → \( \cos 90^\circ = 0 \) → \( W = 0 \)
- \( \theta = 180^\circ \): Force opposite to motion → \( \cos 180^\circ = -1 \) → negative work (removes energy)
Example:
A worker pulls a crate across a floor with a force of \( 80 \, \text{N} \) at an angle of \( 30^\circ \) to the horizontal. The crate moves \( 5.0 \, \text{m} \) horizontally. Calculate the work done by the force.
▶️ Answer/Explanation
Step 1: Use the work formula
\( W = F s \cos \theta \)
\( = 80 \cdot 5.0 \cdot \cos(30^\circ) \)
Step 2: Calculate
\( \cos(30^\circ) = 0.866 \)
\( W = 80 \cdot 5 \cdot 0.866 = \boxed{346.4 \, \text{J}} \)
Interpretation:
Only the horizontal component of the applied force contributes to the work done in moving the crate. The vertical component does not contribute, as there’s no vertical displacement.
Work Done by the Resultant Force on a System
According to the work-energy principle, the work done by the net (resultant) force on an object is equal to the total change in the mechanical energy of the object:
\( W_{\text{net}} = \Delta E = \Delta K + \Delta U \)
Where:
- \( W_{\text{net}} \): Work done by the net external force
- \( \Delta K \): Change in kinetic energy = \( \frac{1}{2}mv^2 – \frac{1}{2}mu^2 \)
- \( \Delta U \): Change in potential energy, if applicable
This principle connects Newton’s second law to energy: a net force causes acceleration, and hence a change in kinetic energy. If the object also moves vertically, gravitational potential energy changes too.
Example:
A 2.0 kg object is pushed from rest by a constant horizontal force across a frictionless surface. After moving 4.0 m, its speed is 6.0 m/s. Calculate the work done by the force and the change in kinetic energy.
▶️ Answer/Explanation
Step 1: Initial and final kinetic energy
Initial KE = \( \frac{1}{2}mv^2 = \frac{1}{2}(2.0)(0)^2 = 0 \, \text{J} \)
Final KE = \( \frac{1}{2}(2.0)(6.0)^2 = 36 \, \text{J} \)
Step 2: Change in energy
\( \Delta K = 36 – 0 = \boxed{36 \, \text{J}} \)
Step 3: Work-Energy connection
Since there’s no potential energy change or friction, the net work done by the force = \( \Delta K = 36 \, \text{J} \)
Hence, work done by the net force is equal to the increase in kinetic energy.