📐 Quick Reference:
KE = ½mv²
PE = mgh
Conservation: KE₁ + PE₁ = KE₂ + PE₂
Work = F × d
Power = Work / time
g = 10 m/s²
Problem 1.1Beginner
A 2 kg ball is moving at 5 m/s. Calculate its kinetic energy.
Step-by-Step Solution:
Given: m = 2 kg, v = 5 m/s
Formula: KE = ½mv²
Calculation: KE = ½ × 2 × (5)²
KE = 1 × 25 = 25 J
✅ Answer: 25 J
Problem 1.2Beginner
A 5 kg book is on a shelf 3 meters high. What is its gravitational potential energy? (g = 10 m/s²)
Step-by-Step Solution:
Given: m = 5 kg, h = 3 m, g = 10 m/s²
Formula: PE = mgh
Calculation: PE = 5 × 10 × 3 = 150 J
✅ Answer: 150 J
🔍 Observation Connection: This is the energy stored in the book. If it falls, this converts to kinetic energy!
Problem 1.3Beginner
A force of 20 N pushes a box 4 meters. How much work is done?
Step-by-Step Solution:
Given: F = 20 N, d = 4 m
Formula: W = F × d
Calculation: W = 20 × 4 = 80 J
✅ Answer: 80 J
Problem 2.1Intermediate
A 500 kg roller coaster car starts from rest at the top of a 40 m hill. Assuming no friction, find its speed at the bottom. (g = 10 m/s²)
🎢 Roller Coaster
Top: h = 40 m, v = 0
│
│
│
│
▼
Bottom: h = 0, v = ?
Energy conservation: mgh = ½mv²
Step-by-Step Solution:
Given: m = 500 kg, h = 40 m, g = 10 m/s²
Conservation of Energy: PE_top = KE_bottom
mgh = ½mv²
Mass cancels out! gh = ½v²
10 × 40 = ½v²
400 = ½v²
v² = 800
v = √800 = 28.3 m/s
✅ Answer: 28.3 m/s
🧠 Notice: Mass canceled! All objects fall at same speed regardless of mass (without friction).
Problem 2.2Intermediate
A 2 kg ball is dropped from a height of 20 m. Find its speed when it hits the ground. (g = 10 m/s²)
Step-by-Step Solution:
Given: m = 2 kg, h = 20 m, g = 10 m/s²
Energy conservation: mgh = ½mv²
2 × 10 × 20 = ½ × 2 × v²
400 = v²
v = 20 m/s
✅ Answer: 20 m/s
Problem 2.3Intermediate
A motor lifts a 50 kg crate to a height of 10 meters in 5 seconds. What is the power output of the motor? (g = 10 m/s²)
Step-by-Step Solution:
Given: m = 50 kg, h = 10 m, t = 5 s, g = 10 m/s²
Work done: W = mgh = 50 × 10 × 10 = 5000 J
Power: P = W/t = 5000/5 = 1000 W
✅ Answer: 1000 W (or 1 kW)
Problem 3.1Advanced
A pendulum bob of mass 2 kg is released from a height of 0.8 m above its lowest point. Find its speed at the lowest point. If it then swings up to a height of 0.6 m on the other side, how much energy was lost to friction?
🔍 Observation Connection: This matches our pendulum observation! Each swing loses energy to air resistance and friction.
Problem 3.2Advanced
A roller coaster car of mass 300 kg enters a vertical loop of radius 10 m. What minimum height above the loop must it start from to just make it through the loop without falling? (Assume no friction, g = 10 m/s²)
Loop-the-Loop
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/ \
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| 10m |
| |
\ /
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At top of loop: need minimum speed v = √(gr)
Step-by-Step Solution:
At top of loop (height = 2r = 20 m):
Minimum speed needed: v = √(gr) = √(10 × 10) = 10 m/s
Energy at top: KE + PE = ½mv² + mg(2r)
= ½ × 300 × 100 + 300 × 10 × 20
= 15,000 + 60,000 = 75,000 J
At start (height H): All energy is PE = mgH
Conservation: mgH = 75,000
300 × 10 × H = 75,000
3,000H = 75,000
H = 25 m
✅ Answer: 25 m above ground (or 5 m above top of loop)
🧠 Think Deeper: The car needs extra height to have enough speed at the top! This is why real roller coasters start higher than the loop.
🧠 Critical Thinking Challenge:
In Problem 3.2, what would happen if the starting height was exactly 20 m (same as the top of loop)? Would the car make it through? Why or why not?
Hint: Think about energy conversion and the speed needed at the top!