Question 1.1: Calculate the work done on the girl by gravity and the normal force.
A girl with mass 60 kg slides 3 m down an inclined plane at 40 degrees.
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Step-by-step solution for: 2 Worksheet: Work, energy and power | Schemes and Mind Maps Acting ...
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Show Answer Key & Explanations
Step-by-step solution for: 2 Worksheet: Work, energy and power | Schemes and Mind Maps Acting ...
Let's solve each part of QUESTION 1 step by step, with clear explanations.
---
## 1.1 A girl with mass 60 kg slides 3 m down an inclined plane (40° to horizontal)
Work done by a force is given by:
> W = F × d × cosθ
where:
- F is the magnitude of the force,
- d is the displacement,
- θ is the angle between the force vector and the displacement vector.
Gravity acts vertically downward. The displacement is along the incline, 3 m at 40° to horizontal.
The component of gravity in the direction of displacement (i.e., parallel to the incline) is:
> F_grav_parallel = mg sin(40°)
So, work done by gravity is:
> W_gravity = (mg sin(40°)) × d
Plug in values:
- m = 60 kg
- g = 9.8 m/s²
- d = 3 m
- sin(40°) ≈ 0.6428
> W_gravity = 60 × 9.8 × 0.6428 × 3
> = 60 × 9.8 × 1.9284
> ≈ 60 × 18.898
> ≈ 1133.9 J
✔ Answer: ~1134 J
---
The normal force acts perpendicular to the surface (i.e., perpendicular to the displacement 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
*(This makes sense — normal force doesn’t contribute to motion along the incline.)*
---
## 1.2 A block (5 kg) is pulled by 60 N at 30° to ground; moves 3.25 m
Gravity acts vertically downward. The displacement is horizontal (since it says “to the ground” and no mention of vertical motion).
So, the angle between gravity (downward) and displacement (horizontal) is 90°.
> W_gravity = F_gravity × d × cos(90°) = 0
✔ Answer: 0 J
*(No vertical displacement → gravity does no work.)*
---
Applied force = 60 N at 30° to horizontal. Displacement is horizontal → so angle between force and displacement is 30°.
> W_applied = F × d × cosθ = 60 × 3.25 × cos(30°)
cos(30°) = √3/2 ≈ 0.8660
> W_applied = 60 × 3.25 × 0.8660
> = 195 × 0.8660
> ≈ 168.87 J
✔ Answer: ~169 J
---
## 1.3 A 1200 kg car pulled 3 m up a 30° incline by 8000 N rope; friction = 20 N
Forces acting on the car:
1. Weight (mg): Downward, magnitude = 1200 × 9.8 = 11760 N
2. Normal force (N): Perpendicular to incline, upward
3. Tension (T): Up the incline, 8000 N
4. Friction (f): Down the incline (opposes motion), 20 N
*Note: Since the car is being pulled UP the incline, friction opposes that motion → acts DOWN the incline.*
FBD Sketch (text description):
```
↑ T = 8000 N
|
/| |
/ | |
/ | |
/___|_____|______ incline (30°)
↓ f = 20 N
↓ mg sin(30°) [component down incline]
← N (normal, ⊥ to incline)
↓ mg cos(30°) [component into incline]
```
*(You’d draw arrows labeled with forces and angles.)*
---
Net work = sum of work done by all forces.
We can calculate work done by each force:
#### i. Work by tension (T):
Direction same as displacement → θ = 0°
> W_T = T × d × cos(0°) = 8000 × 3 × 1 = 24,000 J
#### ii. Work by friction (f):
Opposes motion → θ = 180° → cos(180°) = -1
> W_f = 20 × 3 × (-1) = -60 J
#### iii. Work by gravity:
Component of gravity parallel to incline: mg sin(30°) = 1200×9.8×0.5 = 5880 N, acting down the incline.
Displacement is up the incline → angle = 180°
> W_grav = (mg sin(30°)) × d × cos(180°) = 5880 × 3 × (-1) = -17,640 J
#### iv. Work by normal force:
Perpendicular to displacement → cos(90°) = 0 → 0 J
---
Total net work = W_T + W_f + W_grav + W_N
= 24,000 - 60 - 17,640 + 0 = 6,300 J
✔ Answer: 6300 J
*(Alternatively, you could find net force along incline and multiply by distance:*
Net force = T - f - mg sin(30°) = 8000 - 20 - 5880 = 2100 N
W_net = 2100 × 3 = 6300 J — same result!)
---
## 1.4 Block (5 kg) pulled by 50 N at 30°, 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
> W_net = ΔKE = ½mv² - ½mv₀² = ½mv² (since starts from rest → v₀ = 0)
So we need to compute net work.
Forces doing work:
- Applied force: 50 N at 30° to horizontal → component in direction of motion: 50 cos(30°)
- Friction: 4.2 N opposite to motion → negative work
Displacement = 2.60 m
---
> W_app = F_app × d × cos(30°) = 50 × 2.60 × (√3/2)
> = 130 × 0.8660 ≈ 112.58 J
> W_friction = -f × d = -4.2 × 2.60 = -10.92 J
> W_net = 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
---
## 1.5 Film star + car (720 kg) drives 4.8 m down 42° incline; engine force = 200 N, friction = 23 N, starts from rest
Engine force = 200 N. Assuming this force is applied in the direction of motion (down the incline), then:
> W_engine = F_engine × d × cos(0°) = 200 × 4.8 × 1 = 960 J
✔ Answer: 960 J
*(Note: If the engine were pulling uphill, it might be different, but since car is going DOWN and engine is "exerting force", context implies it’s aiding motion → same direction.)*
---
Use work-energy principle again:
> W_net = ΔKE = ½mv²
We need net work done on the car.
Forces doing work:
1. Gravity: Component down incline = mg sin(42°)
2. Engine: 200 N down incline → positive work
3. Friction: 23 N up incline (opposing motion) → negative work
Displacement = 4.8 m down incline.
---
> W_grav = (mg sin(42°)) × d
> m = 720 kg, g = 9.8, sin(42°) ≈ 0.6691
> W_grav = 720 × 9.8 × 0.6691 × 4.8
> First: 720 × 9.8 = 7056
> 7056 × 0.6691 ≈ 4721.5
> 4721.5 × 4.8 ≈ 22,663.2 J
> W_engine = 200 × 4.8 = 960 J
> W_friction = -23 × 4.8 = -110.4 J
> W_net = 22,663.2 + 960 - 110.4 = 23,512.8 J
> ½mv² = 23,512.8
> ½ × 720 × v² = 23,512.8
> 360 v² = 23,512.8
> v² = 23,512.8 / 360 ≈ 65.313
> v = √65.313 ≈ 8.08 m/s
✔ Answer: ~8.08 m/s
---
## ✔ FINAL ANSWERS SUMMARY:
a. Work by gravity: 1134 J
b. Work by normal force: 0 J
a. Work by gravity: 0 J
b. Work by applied force: 169 J
a. FBD: Weight (down), Normal (⊥ incline), Tension (up incline), Friction (down incline)
b. Net work: 6300 J
Speed after 2.60 m: 6.38 m/s
a. Work by engine: 960 J
b. Velocity after 4.8 m: 8.08 m/s
---
Let me know if you’d like diagrams or further clarification!
---
## 1.1 A girl with mass 60 kg slides 3 m down an inclined plane (40° to horizontal)
a. Work done on the girl by gravity
Work done by a force is given by:
> W = F × d × cosθ
where:
- F is the magnitude of the force,
- d is the displacement,
- θ is the angle between the force vector and the displacement vector.
Gravity acts vertically downward. The displacement is along the incline, 3 m at 40° to horizontal.
The component of gravity in the direction of displacement (i.e., parallel to the incline) is:
> F_grav_parallel = mg sin(40°)
So, work done by gravity is:
> W_gravity = (mg sin(40°)) × d
Plug in values:
- m = 60 kg
- g = 9.8 m/s²
- d = 3 m
- sin(40°) ≈ 0.6428
> W_gravity = 60 × 9.8 × 0.6428 × 3
> = 60 × 9.8 × 1.9284
> ≈ 60 × 18.898
> ≈ 1133.9 J
✔ Answer: ~1134 J
---
b. Work done on the girl by the normal force
The normal force acts perpendicular to the surface (i.e., perpendicular to the displacement 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
*(This makes sense — normal force doesn’t contribute to motion along the incline.)*
---
## 1.2 A block (5 kg) is pulled by 60 N at 30° to ground; moves 3.25 m
a. Work done on the block by gravity
Gravity acts vertically downward. The displacement is horizontal (since it says “to the ground” and no mention of vertical motion).
So, the angle between gravity (downward) and displacement (horizontal) is 90°.
> W_gravity = F_gravity × d × cos(90°) = 0
✔ Answer: 0 J
*(No vertical displacement → gravity does no work.)*
---
b. Work done on the block by the applied force
Applied force = 60 N at 30° to horizontal. Displacement is horizontal → so angle between force and displacement is 30°.
> W_applied = F × d × cosθ = 60 × 3.25 × cos(30°)
cos(30°) = √3/2 ≈ 0.8660
> W_applied = 60 × 3.25 × 0.8660
> = 195 × 0.8660
> ≈ 168.87 J
✔ Answer: ~169 J
---
## 1.3 A 1200 kg car pulled 3 m up a 30° incline by 8000 N rope; friction = 20 N
a. Free body diagram (FBD)
Forces acting on the car:
1. Weight (mg): Downward, magnitude = 1200 × 9.8 = 11760 N
2. Normal force (N): Perpendicular to incline, upward
3. Tension (T): Up the incline, 8000 N
4. Friction (f): Down the incline (opposes motion), 20 N
*Note: Since the car is being pulled UP the incline, friction opposes that motion → acts DOWN the incline.*
FBD Sketch (text description):
```
↑ T = 8000 N
|
/| |
/ | |
/ | |
/___|_____|______ incline (30°)
↓ f = 20 N
↓ mg sin(30°) [component down incline]
← N (normal, ⊥ to incline)
↓ mg cos(30°) [component into incline]
```
*(You’d draw arrows labeled with forces and angles.)*
---
b. Net work done on the car
Net work = sum of work done by all forces.
We can calculate work done by each force:
#### i. Work by tension (T):
Direction same as displacement → θ = 0°
> W_T = T × d × cos(0°) = 8000 × 3 × 1 = 24,000 J
#### ii. Work by friction (f):
Opposes motion → θ = 180° → cos(180°) = -1
> W_f = 20 × 3 × (-1) = -60 J
#### iii. Work by gravity:
Component of gravity parallel to incline: mg sin(30°) = 1200×9.8×0.5 = 5880 N, acting down the incline.
Displacement is up the incline → angle = 180°
> W_grav = (mg sin(30°)) × d × cos(180°) = 5880 × 3 × (-1) = -17,640 J
#### iv. Work by normal force:
Perpendicular to displacement → cos(90°) = 0 → 0 J
---
Total net work = W_T + W_f + W_grav + W_N
= 24,000 - 60 - 17,640 + 0 = 6,300 J
✔ Answer: 6300 J
*(Alternatively, you could find net force along incline and multiply by distance:*
Net force = T - f - mg sin(30°) = 8000 - 20 - 5880 = 2100 N
W_net = 2100 × 3 = 6300 J — same result!)
---
## 1.4 Block (5 kg) pulled by 50 N at 30°, 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
> W_net = ΔKE = ½mv² - ½mv₀² = ½mv² (since starts from rest → v₀ = 0)
So we need to compute net work.
Forces doing work:
- Applied force: 50 N at 30° to horizontal → component in direction of motion: 50 cos(30°)
- Friction: 4.2 N opposite to motion → negative work
Displacement = 2.60 m
---
Step 1: Work by applied force
> W_app = F_app × d × cos(30°) = 50 × 2.60 × (√3/2)
> = 130 × 0.8660 ≈ 112.58 J
Step 2: Work by friction
> W_friction = -f × d = -4.2 × 2.60 = -10.92 J
Step 3: Net work
> W_net = 112.58 - 10.92 = 101.66 J
Step 4: 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
---
## 1.5 Film star + car (720 kg) drives 4.8 m down 42° incline; engine force = 200 N, friction = 23 N, starts from rest
a. Work done on the car by the engine
Engine force = 200 N. Assuming this force is applied in the direction of motion (down the incline), then:
> W_engine = F_engine × d × cos(0°) = 200 × 4.8 × 1 = 960 J
✔ Answer: 960 J
*(Note: If the engine were pulling uphill, it might be different, but since car is going DOWN and engine is "exerting force", context implies it’s aiding motion → same direction.)*
---
b. Velocity after moving 4.8 m
Use work-energy principle again:
> W_net = ΔKE = ½mv²
We need net work done on the car.
Forces doing work:
1. Gravity: Component down incline = mg sin(42°)
2. Engine: 200 N down incline → positive work
3. Friction: 23 N up incline (opposing motion) → negative work
Displacement = 4.8 m down incline.
---
Step 1: Work by gravity
> W_grav = (mg sin(42°)) × d
> m = 720 kg, g = 9.8, sin(42°) ≈ 0.6691
> W_grav = 720 × 9.8 × 0.6691 × 4.8
> First: 720 × 9.8 = 7056
> 7056 × 0.6691 ≈ 4721.5
> 4721.5 × 4.8 ≈ 22,663.2 J
Step 2: Work by engine
> W_engine = 200 × 4.8 = 960 J
Step 3: Work by friction
> W_friction = -23 × 4.8 = -110.4 J
Step 4: Net work
> W_net = 22,663.2 + 960 - 110.4 = 23,512.8 J
Step 5: Solve for velocity
> ½mv² = 23,512.8
> ½ × 720 × v² = 23,512.8
> 360 v² = 23,512.8
> v² = 23,512.8 / 360 ≈ 65.313
> v = √65.313 ≈ 8.08 m/s
✔ Answer: ~8.08 m/s
---
## ✔ FINAL ANSWERS SUMMARY:
1.1
a. Work by gravity: 1134 J
b. Work by normal force: 0 J
1.2
a. Work by gravity: 0 J
b. Work by applied force: 169 J
1.3
a. FBD: Weight (down), Normal (⊥ incline), Tension (up incline), Friction (down incline)
b. Net work: 6300 J
1.4
Speed after 2.60 m: 6.38 m/s
1.5
a. Work by engine: 960 J
b. Velocity after 4.8 m: 8.08 m/s
---
Let me know if you’d like diagrams or further clarification!
Parent Tip: Review the logic above to help your child master the concept of calculating work and power worksheet.
