Conservation of Energy — Practice · Grade 12 Science
Chapter: Work, Energy & MomentumWhat this practice covers
By the end of Conservation of Energy you should be able to:
- I can apply conservation of mechanical energy (KE + PE = constant) to solve for speeds and heights in ideal systems.
- I can account for energy dissipated by friction and calculate the efficiency of a real device.
9 questions, numeric, multiple choice and written answer, each with a worked explanation.
Vocabulary used: mechanical energy, conservation of mechanical energy, law of conservation of energy, dissipation, efficiency.
Problem 1
Answer and explanation
120 J
With no friction, mechanical energy is conserved, so all 120 J of PE becomes kinetic energy at the bottom.
Problem 2
Answer and explanation
10 m/s
All 200 J becomes kinetic energy: 200 = ½ × 4.0 × v² = 2v², so v² = 100 and v = 10 m/s.
Problem 3
Answer and explanation
70 %
Efficiency = (350 ÷ 500) × 100% = 70%.
Problem 4
- Twice as tall as the first hill
- Lower than the first hill
- Taller than the first hill
- Exactly the same height as the first hill
Answer and explanation
Lower than the first hill
Friction dissipates some mechanical energy as heat and sound on the way, so the car reaches the second hill with less energy and cannot climb as high — the second hill must be lower.
Problem 5
Answer and explanation
As the pendulum swings, energy trades between gravitational potential energy at the high points and kinetic energy at the bottom, with the mechanical total staying nearly constant. A real pendulum stops because friction at the pivot and air resistance gradually dissipate that mechanical energy into thermal energy and sound; the total energy is still conserved, just no longer in the form of visible swinging motion.
Problem 6
Hints
Mass cancels in this ideal gravitational-energy calculation.
Answer and explanation
65 J under the stated accounting.
Problem 7
Hints
State what the supplied observations or conditions establish, and separate that from any assumption.
Answer and explanation
No. Energy or kinematics gives the same speed for the same initial conditions and g.
Problem 8
Answer and explanation
40 J, because net work is+30 J.
The work–energy theorem states Wnet=ΔK in this ordinary mechanics model. It concerns the sum of all forces’ work on the selected object, not necessarily total energy or one applied force’s contribution. A driving force doing 50 J while resistance does−20 J yields a 30 J kinetic increase. If gravity is among the forces, its work enters this net-work account; a potential-energy formulation reorganizes that term instead of adding it twice.
40 J, because net work is+30 J.
Response criteria
- Interpret labels, structures, axes, units and directions correctly.
- Explain the physical or environmental relationship or mechanism.
- Respect stated assumptions and distinguish observation from inference.
Problem 9
Answer and explanation
The accounted final/output total is 190 J, leaving 10 J. Check boundary heat transfer, deformation/storage and instrument uncertainty rather than claiming energy was destroyed.
The accounted final/output total is 190 J, leaving 10 J.
Check boundary heat transfer, deformation/storage and instrument uncertainty rather than claiming energy was destroyed.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.