Free Collection of Work, Power, and Energy Worksheets - Free Printable

Let’s solve each part step by step. We’ll use the formula for work:

> Work = Force × Distance × cos(θ)
> where θ is the angle between the force and the direction of motion.

Also, remember:
- Gravity acts straight down.
- Normal force is perpendicular to the surface.
- Friction opposes motion.
- Work-energy principle: Net work = change in kinetic energy = ½mv² - ½mu² (if starting from rest, u=0).

---

1.1 Girl sliding down incline

Mass = 60 kg, distance along slope = 3 m, angle = 40°

#### a. Work done by gravity

Gravity pulls straight down. The component of gravity *along* the slope is mg sin(θ). But actually, we can think of it as: the vertical height she drops is h = 3 × sin(40°). Then work by gravity = mgh.

Let’s calculate that:

h = 3 × sin(40°) ≈ 3 × 0.6428 ≈ 1.928 m

Work by gravity = mgh = 60 × 9.8 × 1.928 ≈ 60 × 9.8 × 1.928

First: 9.8 × 1.928 ≈ 18.8944
Then: 60 × 18.8944 ≈ 1133.66 J

Alternatively, using force along slope: F_gravity_parallel = mg sin(40°) = 60×9.8×sin(40°) ≈ 588 × 0.6428 ≈ 378.0 N
Work = F × d = 378.0 × 3 = 1134 J → same thing.

✔ So, work by gravity ≈ 1134 J

#### b. Work done by normal force

Normal force is perpendicular to the slope. Motion is along the slope. So angle between normal force and displacement = 90°.

cos(90°) = 0 → Work = 0

✔ Answer: 0 J

---

1.2 Block pulled at 30°

Mass = 5 kg, force = 60 N at 30° to ground, moves 3.25 m horizontally.

#### a. Work done by gravity

Gravity acts downward. Motion is horizontal. Angle between gravity and motion = 90° → cos(90°)=0 → Work = 0

✔ Answer: 0 J

#### b. Work done by applied force

Force = 60 N, angle with horizontal = 30°, distance = 3.25 m

Work = F × d × cos(θ) = 60 × 3.25 × cos(30°)

cos(30°) = √3/2 ≈ 0.8660

So: 60 × 3.25 = 195
195 × 0.8660 ≈ 168.87 J

✔ Answer: ≈ 168.9 J

---

1.3 Car pulled up incline

Mass = 1200 kg, pulled 3 m up 30° incline, rope force = 8000 N (up the slope), friction = 20 N (down the slope)

#### a. Free body diagram (description since we can’t draw)

Forces on car:

- Weight (mg) straight down → split into components:
- Parallel to slope: mg sin(30°) down the slope
- Perpendicular: mg cos(30°) into slope
- Normal force: perpendicular out from slope
- Applied force (rope): 8000 N up the slope
- Friction: 20 N down the slope (opposing motion)

#### b. Net work done on the car

Net work = sum of work done by all forces.

We only care about forces parallel to motion (since perpendicular forces do zero work).

Motion is up the slope.

Forces along slope:

- Rope: +8000 N (same direction as motion)
- Friction: -20 N (opposite)
- Gravity component: mg sin(30°) down slope → so negative if motion is up

mg = 1200 × 9.8 = 11760 N
mg sin(30°) = 11760 × 0.5 = 5880 N → down slope → so work by gravity = -5880 × 3

Now compute work by each:

- Rope: 8000 × 3 = 24000 J
- Friction: -20 × 3 = -60 J
- Gravity: -5880 × 3 = -17640 J

Total net work = 24000 - 60 - 17640 = ?

24000 - 17640 = 6360
6360 - 60 = 6300 J

✔ Answer: 6300 J

---

1.4 Block pulled with friction — find speed

Mass = 5 kg, force = 50 N at 30°, friction = 4.2 N, distance = 2.60 m, starts from rest.

Use work-energy principle: Net work = ΔKE = ½mv² - 0

First, find net work.

Only horizontal component of applied force does work (vertical component doesn't move vertically? Actually, motion is horizontal, so yes — only horizontal component matters).

Applied force horizontal component: 50 × cos(30°) ≈ 50 × 0.8660 ≈ 43.3 N

Friction opposes: 4.2 N

So net force along motion = 43.3 - 4.2 = 39.1 N

Net work = net force × distance = 39.1 × 2.60 ≈ ?

39.1 × 2.6 = let's compute:

39 × 2.6 = 101.4
0.1 × 2.6 = 0.26
Total = 101.66 J

So net work ≈ 101.66 J = ½ × 5 × v²

→ 101.66 = 2.5 v²
→ v² = 101.66 / 2.5 = 40.664
→ v = √40.664 ≈ 6.377 m/s

✔ Answer: ≈ 6.38 m/s

---

1.5 Film star in car going down incline

Total mass = 720 kg, distance = 4.8 m down 42° incline, starts from rest, friction = 23 N, engine force = 200 N (assumed acting down the slope? Since it says “engine exerts a force” and car is moving down — likely helping motion, so same direction as motion).

Assume engine force is directed down the slope (helping motion).

#### a. Work done by engine

Engine force = 200 N, direction same as motion (down slope), distance = 4.8 m

Work = 200 × 4.8 = 960 J

✔ Answer: 960 J

#### b. Velocity after 4.8 m

Use work-energy principle: Net work = ΔKE = ½mv²

Find net work.

Forces doing work (parallel to motion):

- Engine: +200 N (down slope)
- Friction: -23 N (up slope, opposing motion)
- Gravity component: mg sin(42°) down slope → positive

mg = 720 × 9.8 = 7056 N
sin(42°) ≈ 0.6691
Gravity component = 7056 × 0.6691 ≈ let’s compute:

7000 × 0.6691 = 4683.7
56 × 0.6691 ≈ 37.47
Total ≈ 4721.17 N

So total force down slope: engine + gravity - friction = 200 + 4721.17 - 23 = 4898.17 N

Wait — no! For work, we compute work by each force separately.

Better:

Work by engine = 200 × 4.8 = 960 J
Work by friction = -23 × 4.8 = -110.4 J
Work by gravity = mg sin(42°) × 4.8 = 4721.17 × 4.8 ≈ ?

4721.17 × 4.8:

First, 4700 × 4.8 = 22560
21.17 × 4.8 ≈ 101.616
Total ≈ 22661.616 J

So total net work = 960 - 110.4 + 22661.616 ≈ ?

960 - 110.4 = 849.6
849.6 + 22661.616 = 23511.216 J

This equals ½mv² = 0.5 × 720 × v² = 360 v²

So:

360 v² = 23511.216
v² = 23511.216 / 360 ≈ 65.3089
v = √65.3089 ≈ 8.08 m/s

✔ Answer: ≈ 8.08 m/s

---

## Final Answers:

1.1a: 1134 J
1.1b: 0 J
1.2a: 0 J
1.2b: 168.9 J
1.3b: 6300 J
1.4: 6.38 m/s
1.5a: 960 J
1.5b: 8.08 m/s