CIE IGCSE Physics (0625) Work Study Notes - New Syllabus
CIE IGCSE Physics (0625) Topic 1.7.2 Work Study Notes
Key Concepts:
Core
- Understand that mechanical or electrical work done is equal to the energy transferred
- Recall and use the equation for mechanical working W = Fd = ΔE
Supplement
- No additional Supplement content
Mechanical/Electrical Work = Energy Transferred
Mechanical/Electrical Work = Energy Transferred:
- Whenever mechanical work (done by forces) or electrical work (done by electric currents) is done, it results in a transfer of energy.
- So, work done = energy transferred (in joules).
Therefore:
\( \text{Work done} = \text{Energy transferred} \)
- This applies to:
- Objects moved by a force (mechanical work)
- Components powered by electric current (electrical work)
Example:
An electric heater runs for 10 seconds with a power of 500 W. How much energy is transferred?
▶️ Answer/Explanation
Step 1: Use the relationship
Energy transferred = Power × Time
\( E = P \times t \)
Step 2: Substitute values
\( E = 500 \times 10 = \boxed{5000 \, \text{J}} \)
So, 5000 J of electrical energy is transferred into thermal energy by the heater — this is also the work done.
Example:
A person lifts a 10 kg box vertically to a height of 2 meters. How much work is done? (Take \( g = 9.8 \, \text{m/s}^2 \))
▶️ Answer/Explanation
Step 1: Use the formula for work done (mechanical)
\( W = F \times d \)
Where:
\( F = \text{weight} = mg = 10 \times 9.8 = 98 \, \text{N} \)
\( d = 2 \, \text{m} \)
Step 2: Calculate work
\( W = 98 \times 2 = \boxed{196 \, \text{J}} \)
Conclusion:
196 J of mechanical work is done to lift the box, and this equals the gravitational potential energy gained.
Mechanical Work
Mechanical Work:
- When a force causes an object to move, mechanical work is done.
- The energy transferred is equal to the work done.
Equation for Mechanical Work:
\( W = F \times d = \Delta E \)
- \( W \) = Work done (in joules, J)
- \( F \) = Force applied (in newtons, N)
- \( d \) = Distance moved in the direction of the force (in meters, m)
- \( \Delta E \) = Energy transferred (in joules, J)
Key Point: This equation is valid only when the force is applied in the direction of the movement.
Example:
A child pulls a sledge with a horizontal force of 80 N for a distance of 15 m. Calculate the work done by the child. Assume the force is applied completely in the direction of motion.
▶️ Answer/Explanation
Step 1: Use the work equation
\( W = F \times d \)
Step 2: Substitute values
\( W = 80 \times 15 = \boxed{1200 \, \text{J}} \)
Conclusion:
The child does 1200 J of mechanical work in pulling the sledge. This is also the energy transferred to overcome friction and increase kinetic energy.
Example:
A worker pulls a 20 kg box up a ramp that is 5 m long and inclined at 30°. The pulling force is constant and acts along the ramp. Calculate:
- (a) The work done by the worker
- (b) The gain in gravitational potential energy of the box
- (c) The energy lost to friction if the total work done is 600 J
▶️ Answer/Explanation
Given:
- Mass, \( m = 20 \, \text{kg} \)
- Distance along ramp, \( d = 5 \, \text{m} \)
- Angle of incline, \( \theta = 30^\circ \)
- Gravitational field strength, \( g = 9.8 \, \text{m/s}^2 \)
(a) Work Done by the Worker:
Let’s assume the pulling force equals the component of weight along the ramp:
\( F = mg \sin\theta = 20 \times 9.8 \times \sin(30^\circ) = 196 \, \text{N} \)
\( W = F \times d = 196 \times 5 = \boxed{980 \, \text{J}} \)
(b) Gain in GPE:
Vertical height = \( h = d \sin\theta = 5 \times \sin(30^\circ) = 2.5 \, \text{m} \)
\( \Delta GPE = mgh = 20 \times 9.8 \times 2.5 = \boxed{490 \, \text{J}} \)
(c) Energy Lost to Friction:
Total work done = 600 J (given)
Useful energy (GPE gain) = 490 J
Energy lost = \( 600 – 490 = \boxed{110 \, \text{J}} \)
Conclusion:
Not all the work done went into increasing gravitational potential energy. Some was lost to friction between the box and ramp surface.