Energy & Its Transformations — Science
- Grade 6 — Forms of Energy
- Grade 6 — Energy Transformations & Conservation
- Grade 6 — Heat vs Temperature
- Grade 6 — Heat Transfer: Conduction, Convection & Radiation
- Grade 8 — Kinetic & Potential Energy
↑ Deepens: energy becomes kinetic + potential you can quantify.
- Grade 8 — Energy Transfer & Transformation
- Grade 8 — Thermal Energy & Heat Transfer
- Grade 8 — Conservation of Energy
- Grade 12 — Work & Power
↑ Deepens: energy tied to work, and conserved in collisions.
- Grade 12 — Kinetic & Potential Energy
- Grade 12 — Conservation of Energy
- Grade 12 — Momentum & Collisions
Apply it in a new setting
Problem 1
Hints
Find the kinetic energy at both speeds separately, then subtract.
KE depends on v^2, so doubling the speed quadruples the energy — the two states are not close.
KE lost = ½m(v_0^2 − v^2) = ½ · 1500 · (400 − 100).
Answer and explanation
225000 J
Start: KE = ½ · 1500 · 20^2 = 750 · 400 = 300000 J.
End: KE = ½ · 1500 · 10^2 = 750 · 100 = 75000 J.
Energy lost = 300000 − 75000 = 225000 J.
Problem 2
Hints
At steady speed the lifting force just balances the beam's weight.
Weight = mg, and work = force × distance lifted.
W = (200 · 9.8) · 15.
Answer and explanation
29400 J
Weight of the beam = mg = 200 · 9.8 = 1960 N.
The crane must pull up with 1960 N over 15 m.
W = Fd = 1960 · 15 = 29400 J.
Problem 3
Hints
Power is the work done divided by the time taken.
The work here is the energy given to the load: W = mgh.
P = (50 · 9.8 · 12) ÷ 8.
Answer and explanation
735 W
Work done lifting the load = mgh = 50 · 9.8 · 12 = 5880 J.
Power = work ÷ time = 5880 ÷ 8.
P = 735 W.
Problem 4
Hints
All the gravitational potential energy at the top becomes kinetic energy at the bottom.
Set mgh = ½mv^2 — the mass cancels out, so you don't even need it.
v = √(2gh) = √(2 · 9.8 · 8).
Answer and explanation
12.5 m/s
Conservation of energy: mgh = ½mv^2, so the mass cancels.
v = √(2gh) = √(2 · 9.8 · 8) = √156.8.
v ≈ 12.5 m/s.
Problem 5
Hints
ΔT is the temperature change, not the final temperature.
ΔT = 80 − 20 = 60°C.
q = 2 · 4184 · 60.
Answer and explanation
502080 J
Temperature change ΔT = 80 − 20 = 60°C.
q = mcΔT = 2 · 4184 · 60.
q = 502080 J (about 502 kJ).
Problem 6
Hints
Total momentum before the collision equals total momentum after.
Momentum before = (2 · 3) + (1 · 0); after, the moving mass is 2 + 1 = 3 kg.
v = total momentum ÷ combined mass = 6 ÷ 3.
Answer and explanation
2 m/s
Momentum before = 2 · 3 + 1 · 0 = 6 kg·m/s.
After sticking, combined mass = 2 + 1 = 3 kg, so 3 · v = 6.
v = 6 ÷ 3 = 2 m/s.
Problem 7
- Elastic potential energy → kinetic energy → gravitational potential energy
- Kinetic energy → elastic potential energy → gravitational potential energy
- Gravitational potential energy → kinetic energy → elastic potential energy
- Thermal energy → kinetic energy → elastic potential energy
Hints
Think about where the energy is stored at the very start, while the band is stretched.
At the highest point the stone is momentarily slow but high up — which store is full there?
Stretched band (elastic) → moving stone (kinetic) → high stone (gravitational).
Answer and explanation
Elastic potential energy → kinetic energy → gravitational potential energy
The stretched elastic band stores elastic potential energy at release.
As it snaps back, the stone speeds up — that is kinetic energy.
Rising to the top, the stone slows while gaining height, storing gravitational potential energy, so the order is A.