Free Collection of Work, Power, and Energy Worksheets - Free Printable
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Step-by-step solution for: Free Collection of Work, Power, and Energy Worksheets
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Step-by-step solution for: Free Collection of Work, Power, and Energy Worksheets
Here is the complete solution to Worksheet 2: Work, Energy and Power, covering all parts of Questions 1.1 to 1.5.
---
## QUESTION 1.1
> A girl with mass 60 kg slides 3 m down an inclined plane at 40° to the horizontal.
Concept:
Work done by gravity = force of gravity × displacement in the direction of the force.
Gravity acts vertically downward. The component of displacement in the vertical direction is:
> `h = d × sin(θ)`
where `d = 3 m`, `θ = 40°`
So,
> `h = 3 × sin(40°) ≈ 3 × 0.6428 ≈ 1.928 m`
Force of gravity (weight) = `mg = 60 × 9.8 = 588 N`
Work done by gravity:
> `W_gravity = F_g × h = 588 × 1.928 ≈ 1133.7 J`
✔ Answer: ~1134 J (or 1130 J if rounded to 3 significant figures)
---
Concept:
The normal force is perpendicular to the surface of the incline. The displacement is along the incline. Since the angle between the normal force and displacement is 90°, and `cos(90°) = 0`:
> `W_normal = F_normal × d × cos(90°) = 0`
✔ Answer: 0 J
---
## QUESTION 1.2
> A block (5 kg) is pulled by a force of 60 N at 30° to the ground. It moves 3.25 m.
Concept:
Gravity acts vertically downward. The displacement is horizontal (along the ground). So, the vertical component of displacement is zero → no work done by gravity.
> `W_gravity = 0 J`
✔ Answer: 0 J
---
Concept:
Work = Force × displacement × cos(θ), where θ is the angle between force and displacement.
> `F = 60 N`, `d = 3.25 m`, `θ = 30°`
> `W_applied = 60 × 3.25 × cos(30°) = 60 × 3.25 × (√3/2) ≈ 60 × 3.25 × 0.8660`
> `≈ 60 × 2.8145 ≈ 168.87 J`
✔ Answer: ~169 J
---
## QUESTION 1.3
> A 1200 kg car is pulled 3 m up an incline (30°) by a rope exerting 8000 N. Frictional force = 20 N.
Draw the car on the incline (angle 30°). Forces acting:
1. Weight (mg): Vertically downward → magnitude = 1200 × 9.8 = 11760 N
- Resolve into components:
- Parallel to incline (downhill): `mg sin(30°) = 11760 × 0.5 = 5880 N`
- Perpendicular to incline (into plane): `mg cos(30°) = 11760 × √3/2 ≈ 10184 N`
2. Normal force (N): Perpendicular to incline, upward → balances `mg cos(30°)` → so `N = 10184 N`
3. Tension (T): Up the incline → 8000 N
4. Friction (f): Down the incline (opposes motion) → 20 N
Label all forces clearly with arrows and magnitudes.
*(Since we can't draw here, describe as above — this is what you should sketch.)*
---
Concept:
Net work = sum of work done by all forces.
- Work by tension: `W_T = T × d × cos(0°) = 8000 × 3 × 1 = 24000 J` (up incline)
- Work by friction: `W_f = f × d × cos(180°) = 20 × 3 × (-1) = -60 J`
- Work by gravity (parallel component): `W_g = mg sin(30°) × d × cos(180°) = 5880 × 3 × (-1) = -17640 J`
- Work by normal force: 0 J (perpendicular to displacement)
> Net work = W_T + W_f + W_g + W_N = 24000 - 60 - 17640 + 0 = 6300 J
✔ Answer: 6300 J
---
## QUESTION 1.4
> A 5 kg block is pulled by 50 N at 30° to the ground. Friction = 4.2 N. Moves 2.60 m from rest. Use work-energy principle to find speed.
Work-Energy Principle:
Net work done = change in kinetic energy = `½mv² - ½mv₀²`
Initial velocity = 0 → `ΔKE = ½mv²`
Forces doing work:
- Applied force: `W_app = F × d × cos(30°) = 50 × 2.60 × cos(30°) = 50 × 2.60 × 0.8660 ≈ 112.58 J`
- Friction: `W_friction = f × d × cos(180°) = 4.2 × 2.60 × (-1) = -10.92 J`
- Gravity & Normal force: No vertical displacement → 0 work
> Net work = 112.58 - 10.92 = 101.66 J
> `½mv² = 101.66`
> `½ × 5 × v² = 101.66`
> `2.5 v² = 101.66`
> `v² = 101.66 / 2.5 = 40.664`
> `v = √40.664 ≈ 6.38 m/s`
✔ Answer: ~6.38 m/s
---
## QUESTION 1.5
> Film star + car = 720 kg, moves 4.8 m down a 42° incline. Starts from rest. Friction = 23 N. Engine exerts 200 N *down* the incline.
Engine force = 200 N down the incline. Displacement = 4.8 m down the incline → same direction.
> `W_engine = F × d × cos(0°) = 200 × 4.8 × 1 = 960 J`
✔ Answer: 960 J
---
Use work-energy theorem: Net work = ΔKE = ½mv²
#### Step 1: Find all works
- Work by engine: +960 J (as above)
- Work by gravity: Component of gravity down incline = `mg sin(42°)`
> `F_g_parallel = 720 × 9.8 × sin(42°) ≈ 720 × 9.8 × 0.6691 ≈ 4785.6 N`
> `W_gravity = 4785.6 × 4.8 × cos(0°) ≈ 22971 J` (down incline → positive)
- Work by friction: Opposes motion → up incline → angle = 180°
> `W_friction = 23 × 4.8 × (-1) = -110.4 J`
- Normal force: 0 work
#### Step 2: Total net work
> `W_net = 960 + 22971 - 110.4 = 23820.6 J`
#### Step 3: Set equal to KE
> `½mv² = 23820.6`
> `½ × 720 × v² = 23820.6`
> `360 v² = 23820.6`
> `v² = 23820.6 / 360 ≈ 66.168`
> `v = √66.168 ≈ 8.13 m/s`
✔ Answer: ~8.13 m/s
---
## ✔ FINAL SUMMARY OF ANSWERS:
a. 1134 J
b. 0 J
a. 0 J
b. 169 J
a. *(FBD described above)*
b. 6300 J
6.38 m/s
a. 960 J
b. 8.13 m/s
---
Let me know if you’d like diagrams or step-by-step calculations for any part!
---
## QUESTION 1.1
> A girl with mass 60 kg slides 3 m down an inclined plane at 40° to the horizontal.
a. Work done on the girl by gravity
Concept:
Work done by gravity = force of gravity × displacement in the direction of the force.
Gravity acts vertically downward. The component of displacement in the vertical direction is:
> `h = d × sin(θ)`
where `d = 3 m`, `θ = 40°`
So,
> `h = 3 × sin(40°) ≈ 3 × 0.6428 ≈ 1.928 m`
Force of gravity (weight) = `mg = 60 × 9.8 = 588 N`
Work done by gravity:
> `W_gravity = F_g × h = 588 × 1.928 ≈ 1133.7 J`
✔ Answer: ~1134 J (or 1130 J if rounded to 3 significant figures)
---
b. Work done on the girl by the normal force
Concept:
The normal force is perpendicular to the surface of the incline. The displacement is along the incline. Since the angle between the normal force and displacement is 90°, and `cos(90°) = 0`:
> `W_normal = F_normal × d × cos(90°) = 0`
✔ Answer: 0 J
---
## QUESTION 1.2
> A block (5 kg) is pulled by a force of 60 N at 30° to the ground. It moves 3.25 m.
a. Work done on the block by gravity
Concept:
Gravity acts vertically downward. The displacement is horizontal (along the ground). So, the vertical component of displacement is zero → no work done by gravity.
> `W_gravity = 0 J`
✔ Answer: 0 J
---
b. Work done on the block by the applied force
Concept:
Work = Force × displacement × cos(θ), where θ is the angle between force and displacement.
> `F = 60 N`, `d = 3.25 m`, `θ = 30°`
> `W_applied = 60 × 3.25 × cos(30°) = 60 × 3.25 × (√3/2) ≈ 60 × 3.25 × 0.8660`
> `≈ 60 × 2.8145 ≈ 168.87 J`
✔ Answer: ~169 J
---
## QUESTION 1.3
> A 1200 kg car is pulled 3 m up an incline (30°) by a rope exerting 8000 N. Frictional force = 20 N.
a. Free body diagram (FBD)
Draw the car on the incline (angle 30°). Forces acting:
1. Weight (mg): Vertically downward → magnitude = 1200 × 9.8 = 11760 N
- Resolve into components:
- Parallel to incline (downhill): `mg sin(30°) = 11760 × 0.5 = 5880 N`
- Perpendicular to incline (into plane): `mg cos(30°) = 11760 × √3/2 ≈ 10184 N`
2. Normal force (N): Perpendicular to incline, upward → balances `mg cos(30°)` → so `N = 10184 N`
3. Tension (T): Up the incline → 8000 N
4. Friction (f): Down the incline (opposes motion) → 20 N
Label all forces clearly with arrows and magnitudes.
*(Since we can't draw here, describe as above — this is what you should sketch.)*
---
b. Net work done on the car
Concept:
Net work = sum of work done by all forces.
- Work by tension: `W_T = T × d × cos(0°) = 8000 × 3 × 1 = 24000 J` (up incline)
- Work by friction: `W_f = f × d × cos(180°) = 20 × 3 × (-1) = -60 J`
- Work by gravity (parallel component): `W_g = mg sin(30°) × d × cos(180°) = 5880 × 3 × (-1) = -17640 J`
- Work by normal force: 0 J (perpendicular to displacement)
> Net work = W_T + W_f + W_g + W_N = 24000 - 60 - 17640 + 0 = 6300 J
✔ Answer: 6300 J
---
## QUESTION 1.4
> A 5 kg block is pulled by 50 N at 30° to the ground. Friction = 4.2 N. Moves 2.60 m from rest. Use work-energy principle to find speed.
Work-Energy Principle:
Net work done = change in kinetic energy = `½mv² - ½mv₀²`
Initial velocity = 0 → `ΔKE = ½mv²`
Step 1: Calculate net work
Forces doing work:
- Applied force: `W_app = F × d × cos(30°) = 50 × 2.60 × cos(30°) = 50 × 2.60 × 0.8660 ≈ 112.58 J`
- Friction: `W_friction = f × d × cos(180°) = 4.2 × 2.60 × (-1) = -10.92 J`
- Gravity & Normal force: No vertical displacement → 0 work
> Net work = 112.58 - 10.92 = 101.66 J
Step 2: Set equal to KE
> `½mv² = 101.66`
> `½ × 5 × v² = 101.66`
> `2.5 v² = 101.66`
> `v² = 101.66 / 2.5 = 40.664`
> `v = √40.664 ≈ 6.38 m/s`
✔ Answer: ~6.38 m/s
---
## QUESTION 1.5
> Film star + car = 720 kg, moves 4.8 m down a 42° incline. Starts from rest. Friction = 23 N. Engine exerts 200 N *down* the incline.
a. Work done on the car by the engine
Engine force = 200 N down the incline. Displacement = 4.8 m down the incline → same direction.
> `W_engine = F × d × cos(0°) = 200 × 4.8 × 1 = 960 J`
✔ Answer: 960 J
---
b. Velocity after moving 4.8 m
Use work-energy theorem: Net work = ΔKE = ½mv²
#### Step 1: Find all works
- Work by engine: +960 J (as above)
- Work by gravity: Component of gravity down incline = `mg sin(42°)`
> `F_g_parallel = 720 × 9.8 × sin(42°) ≈ 720 × 9.8 × 0.6691 ≈ 4785.6 N`
> `W_gravity = 4785.6 × 4.8 × cos(0°) ≈ 22971 J` (down incline → positive)
- Work by friction: Opposes motion → up incline → angle = 180°
> `W_friction = 23 × 4.8 × (-1) = -110.4 J`
- Normal force: 0 work
#### Step 2: Total net work
> `W_net = 960 + 22971 - 110.4 = 23820.6 J`
#### Step 3: Set equal to KE
> `½mv² = 23820.6`
> `½ × 720 × v² = 23820.6`
> `360 v² = 23820.6`
> `v² = 23820.6 / 360 ≈ 66.168`
> `v = √66.168 ≈ 8.13 m/s`
✔ Answer: ~8.13 m/s
---
## ✔ FINAL SUMMARY OF ANSWERS:
1.1
a. 1134 J
b. 0 J
1.2
a. 0 J
b. 169 J
1.3
a. *(FBD described above)*
b. 6300 J
1.4
6.38 m/s
1.5
a. 960 J
b. 8.13 m/s
---
Let me know if you’d like diagrams or step-by-step calculations for any part!
Parent Tip: Review the logic above to help your child master the concept of work and energy worksheet.
