Physics practice worksheet focusing on work, energy, and power concepts with various word problems.

Let’s solve each problem one by one, step by step. We’ll use the basic physics formulas for work and energy:

- Work (W) = Force (F) × Distance (d) × cos(θ)
→ If force is in the same direction as motion, θ = 0°, so cos(0°) = 1 → W = F × d

- Gravitational Potential Energy (PE) = mass (m) × gravity (g) × height (h)
→ g ≈ 9.8 m/s² (we’ll use this unless told otherwise)

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Problem 1:
> How far can a mother push a 20.0 kg baby carriage, using a force of 62 N, if she can only do 2920 J of work?

We know:
- Work (W) = 2920 J
- Force (F) = 62 N
- We need to find distance (d)

Formula: W = F × d → d = W / F

d = 2920 ÷ 62 = ?

Let’s calculate:
62 × 47 = 2914 → close!
2920 - 2914 = 6 → so 47 + 6/62 ≈ 47.096...

But let’s do exact division:
2920 ÷ 62 = 47.09677... → round to 3 significant figures? The numbers given are 20.0 (3 sig fig), 62 (2 sig fig), 2920 (3 or 4? ambiguous — but likely 3). So we’ll go with 2 sig figs because 62 has 2.

Wait — actually, 62 might be considered 2 sig figs, but in many school problems, they expect you to use all digits and round at end. Let’s compute exactly:

2920 ÷ 62 = 47.09677419...

But let’s check: 62 × 47 = 2914 → remainder 6 → so 47 + 6/62 = 47 + 3/31 ≈ 47.097 m

In most cases, we’d report as 47.1 m (if rounding to 3 sig figs) or 47 m (if 2 sig figs). But since 2920 has 3 sig figs (assuming it’s not 2920. with decimal), and 62 has 2, the answer should have 2 sig figs → 47 m

Actually, wait — 62 could be exact? In context, probably not. Let’s keep it as 47.1 m for now and see.

But let me recalculate: 62 × 47.1 = 62×47 + 62×0.1 = 2914 + 6.2 = 2920.2 → very close to 2920. So 47.1 m is good.

✔ Final Answer for Problem 1: 47.1 m

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Problem 2:
> Calculate the work done by a 47 N force pushing a pencil 0.26 m.

W = F × d = 47 N × 0.26 m

Calculate: 47 × 0.26

47 × 0.2 = 9.4
47 × 0.06 = 2.82
Total = 9.4 + 2.82 = 12.22 J

Sig figs: 47 (2 sig figs), 0.26 (2 sig figs) → answer should have 2 sig figs → 12 J

But 12.22 rounds to 12? Actually, 12.22 to 2 sig figs is 12. But sometimes they write 12. J to show 2 sig figs. However, in practice, 12 J is fine.

Wait — 47 × 0.26 = let’s do it properly:

47 × 26 = 1222 → then divide by 100 → 12.22 → yes.

So 12 J (with 2 sig figs)

✔ Final Answer for Problem 2: 12 J

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Problem 3:
> A girl pushes her little brother on his sled with a force of 300 N for 750 m. How much work is this if the force of friction acting on the sled is (a) 200 N, (b) 300 N?

Important: Work done BY THE GIRL is based on the force SHE applies, not net force. Friction affects net work or acceleration, but the question says “how much work is this” — referring to the work done by the girl’s push.

So regardless of friction, if she applies 300 N over 750 m, then:

W = F × d = 300 N × 750 m = 225,000 J

Friction doesn’t change how much work *she* does — it changes how much useful work goes into moving the sled vs. heat, etc. But the question is asking for the work done by her force.

So for both (a) and (b), the work done by the girl is the same: 300 × 750 = 225,000 J

But let’s read carefully: “How much work is this if the force of friction...” — maybe they mean net work? But typically, when someone says “work done by a force”, it’s that force times distance.

However, in some contexts, “work done on the sled” might mean net work. But the wording is: “A girl pushes... with a force of 300 N... How much work is this...” — “this” refers to her pushing.

To be safe, let’s consider both interpretations.

If it’s work done by the girl: always 300 N × 750 m = 225,000 J

If it’s net work: then net force = applied force - friction

(a) Net force = 300 - 200 = 100 N → W_net = 100 × 750 = 75,000 J

(b) Net force = 300 - 300 = 0 N → W_net = 0 J

But the problem says “how much work is this” — and “this” likely refers to the work done by the girl’s action. Also, in part (b), if friction equals applied force, the sled moves at constant velocity, but the girl still does work — it just gets dissipated as heat.

I think the intended answer is the work done by the girl, so 225,000 J for both.

But let’s look at similar problems — often they ask for work done by the person, ignoring friction for that calculation.

However, the inclusion of friction suggests they might want net work. Hmm.

Another way: the problem says “how much work is this” — and “this” is the scenario described: pushing with 300 N for 750 m. So I think it’s 300*750.

But to cover bases, perhaps the question is tricking us. Let me see the exact wording: “How much work is this if the force of friction...”

I think it’s poorly worded, but in standard physics, when you’re asked “how much work is done by a force”, it’s that force times distance. Friction is separate.

Moreover, in part (b), if friction is 300 N, and she pushes with 300 N, the sled doesn’t accelerate, but if it’s already moving, she still does work to overcome friction.

So I believe the answer is 225,000 J for both (a) and (b).

But let’s calculate:

300 × 750 = 225,000 J = 2.25 × 10^5 J

Sig figs: 300 (1 or 3? ambiguous — if 300. it’s 3, else 1; but in context, probably 3), 750 (2 or 3? 750 could be 2 sig figs). To be safe, 2.3 × 10^5 J? But 300 and 750 both have at least 2 sig figs? 300 has 1 if no decimal, but in physics problems, often treated as exact or 3.

I think we’ll go with 225,000 J or 2.25 × 10^5 J.

For (a) and (b), same answer.

But I recall that in some textbooks, they specify “work done against friction” or “net work”. Here, it’s not specified.

Given that, and to match common expectations, I’ll assume they want the work done by the applied force, so:

(a) 225,000 J
(b) 225,000 J

But let’s double-check with logic: if friction is present, the girl still exerts 300 N over 750 m, so she does 225,000 J of work. Some of it goes to overcoming friction, some to kinetic energy, but the total work she does is still F_applied * d.

Yes.

✔ Final Answer for Problem 3:
(a) 225,000 J
(b) 225,000 J

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Problem 4:
> A 75.0 kg man pushes on a 500,000 t wall for 250 s but it does not move. How much work does he do on the wall?

Work = Force × Distance × cosθ

If the wall does not move, distance = 0.

So W = F × 0 = 0 J

Time doesn’t matter — work requires displacement.

Even though he’s pushing hard, if there’s no movement, no work is done on the wall.

✔ Final Answer for Problem 4: 0 J

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Problem 5:
> A boy on a bicycle drags a wagon full of newspapers at 0.80 m/s for 30 min using a force of 40 N. How much work has the boy done?

First, find distance traveled.

Speed = 0.80 m/s
Time = 30 minutes = 30 × 60 = 1800 seconds

Distance = speed × time = 0.80 m/s × 1800 s = 1440 m

Now, work = force × distance = 40 N × 1440 m = ?

40 × 1440 = 40 × 1400 = 56,000; 40 × 40 = 1,600; total 57,600 J

Or: 40 × 1440 = 57,600 J

Sig figs: 0.80 (2 sig figs), 30 min (exact? or 1 sig fig? 30 could be 1 or 2; but 30 min is likely exact conversion), 40 N (1 or 2 sig figs? probably 2). So answer should have 2 sig figs → 58,000 J or 5.8 × 10^4 J

57,600 rounded to 2 sig figs is 58,000 J.

But let’s confirm calculation:

0.80 × 1800 = 1440 m (exactly, since 0.80 has two decimals but it’s 8.0×10^{-1}, so 2 sig figs)

40 × 1440 = 57,600

With 2 sig figs: 5.8 × 10^4 J

✔ Final Answer for Problem 5: 58,000 J or 5.8 × 10⁴ J

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Problem 6:
> What is the gravitational potential energy of a 61.2 kg person standing on the roof of a 10-storey building relative to (a) the tenth floor, (b) the sixth floor, (c) the first floor. (Each storey is 2.50 m high.)

Gravitational PE = mgh

m = 61.2 kg
g = 9.8 m/s² (standard value)
h = height above reference point

Each storey = 2.50 m

(a) Relative to tenth floor: the person is on the roof of the 10-storey building. Assuming "roof of 10-storey" means top of 10th floor, so height above 10th floor is 0.

So h = 0 → PE = 0 J

(b) Relative to sixth floor: how many floors above sixth? From 6th to 10th is 4 floors up.

Height = 4 × 2.50 m = 10.0 m

PE = mgh = 61.2 × 9.8 × 10.0

First, 61.2 × 9.8

61.2 × 10 = 612
61.2 × 0.2 = 12.24 → so 612 - 12.24 = wait no: 9.8 = 10 - 0.2, but multiplication: better 61.2 × 9.8

61.2 × 9 = 550.8
61.2 × 0.8 = 48.96
Total = 550.8 + 48.96 = 599.76

Then × 10.0 = 5997.6 J

Sig figs: 61.2 (3), 9.8 (2), 10.0 (3) → limiting is 2 sig figs from g? But g is often taken as exact in such problems, or we use 9.80. Typically in schools, they use g=9.8 and consider it 2 sig figs, but here masses and heights have 3, so perhaps keep 3.

61.2 has 3, 9.8 has 2, 10.0 has 3 → so answer should have 2 sig figs? That would be 6000 J, but that seems rough.

Commonly, g is treated as having more precision. Let’s calculate numerically.

61.2 × 9.8 = 599.76
× 10 = 5997.6 J ≈ 6000 J if 2 sig figs, but 6.00 × 10^3 J for 3 sig figs.

Since 61.2 and 10.0 have 3, and 9.8 is standard, I’ll go with 6000 J but write as 6.00 × 10^3 J? Wait, 5997.6 is closer to 6000.

But let’s do exact: 61.2 * 9.8 * 10 = 61.2 * 98 = ?

61.2 * 100 = 6120
61.2 * 2 = 122.4 → so 6120 - 122.4 = 5997.6? No: 98 = 100 - 2, so 61.2*98 = 61.2*(100-2)=6120 - 122.4=5997.6

Yes.

So 5997.6 J. With sig figs, since 9.8 has 2, but in many curricula, they accept 6000 J or 5998 J. I think for accuracy, we'll report 6000 J, but let's see the other parts.

(c) Relative to first floor: from 1st to 10th is 9 floors up? Or 10 floors?

Building has 10 storeys. Roof of 10-storey building — typically, the roof is above the 10th floor, so height from ground to roof is 10 × 2.50 m = 25.0 m.

Relative to first floor: first floor is at height 0? Or is first floor at ground level?

Usually, "relative to first floor" means height above the first floor level.

If each storey is 2.50 m high, then:

- First floor: height 0 m (reference)
- Second floor: 2.50 m
- ...
- Tenth floor: 9 × 2.50 m? Or 10 × 2.50 m?

This is ambiguous. Typically, the height of the nth floor is (n-1) times storey height if first floor is at ground.

But the problem says: "standing on the roof of a 10-storey building". And "each storey is 2.50 m high".

Usually, a 10-storey building has 10 floors, so the roof is at height 10 × 2.50 m = 25.0 m above ground.

First floor is at ground level, so height above first floor is 25.0 m.

Sixth floor: if first floor is at 0, sixth floor is at 5 × 2.50 = 12.5 m? Or 6 × 2.50?

Standard interpretation: the height of the k-th floor is (k-1) * storey_height, because the first floor is at ground.

But the roof of a 10-storey building is usually at 10 * storey_height.

Let me clarify:

- Ground level: 0 m
- After 1st storey: 2.50 m (top of 1st floor, which is ceiling of 1st floor, floor of 2nd floor)
- So the floor of the nth storey is at (n-1)*2.50 m
- The roof of the building (top of 10th storey) is at 10 * 2.50 = 25.0 m

Now, "relative to the tenth floor": the tenth floor's floor is at 9 * 2.50 = 22.5 m, but the person is on the roof, which is at 25.0 m, so height above tenth floor is 25.0 - 22.5 = 2.5 m? That doesn't make sense with the problem.

The problem says: "standing on the roof of a 10-storey building relative to (a) the tenth floor"

Typically, "the tenth floor" means the level of the tenth floor, which is at height 9*2.50 = 22.5 m if first floor is at 0.

But the roof is at 25.0 m, so for (a) relative to tenth floor, h = 25.0 - 22.5 = 2.5 m

But that seems odd, and the problem might intend that the roof is at the same level as the tenth floor's top, but usually roof is additional.

To simplify, in many problems, they consider the height of the building as number of storeys times storey height, and "relative to nth floor" means height difference.

But let's read: "relative to (a) the tenth floor" — if the person is on the roof of the 10-storey building, and we take the tenth floor as reference, then if the tenth floor is the top floor, its height is 9*2.50 = 22.5 m, roof is 25.0 m, so h=2.5 m.

But that would make (a) not zero, which contradicts my earlier thought.

Perhaps "roof of 10-storey building" means the top of the 10th floor, so height 10*2.50 = 25.0 m, and "tenth floor" means the floor level of the 10th floor, which is at 9*2.50 = 22.5 m, so h=2.5 m for (a).

But the problem says "relative to the tenth floor", and in common parlance, if you're on the roof, you're above the tenth floor.

However, in some interpretations, the "tenth floor" includes up to the roof, but that's not standard.

Another way: perhaps the building has 10 storeys, so there are 10 levels, and the roof is at the top of the 10th storey, so height 10*2.50 = 25.0 m.

The first floor is at 0 m, second at 2.50 m, ..., tenth floor at 9*2.50 = 22.5 m? No.

Let's define:

- The ground floor (first floor) is at height 0 m.
- The second floor is at 2.50 m.
- The third at 5.00 m.
- ...
- The nth floor is at (n-1)*2.50 m.
- The roof of the building is at 10*2.50 = 25.0 m (since there are 10 storeys, each 2.50 m high, so total height 25.0 m).

So:
- Relative to tenth floor: tenth floor is at (10-1)*2.50 = 22.5 m, so h = 25.0 - 22.5 = 2.5 m
- Relative to sixth floor: sixth floor at (6-1)*2.50 = 12.5 m, h = 25.0 - 12.5 = 12.5 m
- Relative to first floor: first floor at 0 m, h = 25.0 m

But the problem says "relative to the tenth floor", and if the person is on the roof, it should be above, but in part (a), if it's 2.5 m, then PE is not zero.

However, in many textbook problems, when they say "on the roof of a 10-storey building" and "relative to the tenth floor", they might mean that the tenth floor is the top, and roof is included, so h=0 for (a).

This is confusing.

Let me look for clues. The problem says: "each storey is 2.50 m high", and "10-storey building", so total height 25.0 m.

"Relative to the tenth floor" — if the tenth floor is the highest occupied floor, its ceiling is at 25.0 m, so if the person is on the roof, which is at 25.0 m, then relative to the tenth floor's ceiling, h=0.

But typically, "floor" refers to the walking surface, not the ceiling.

To resolve this, I think the intended interpretation is:

- The height of the roof is 10 * 2.50 = 25.0 m above ground.
- The first floor is at 0 m.
- The sixth floor is at 5 * 2.50 = 12.5 m (since from 1st to 6th is 5 intervals).
- The tenth floor is at 9 * 2.50 = 22.5 m.

So for (a) relative to tenth floor: h = 25.0 - 22.5 = 2.5 m
(b) relative to sixth floor: h = 25.0 - 12.5 = 12.5 m
(c) relative to first floor: h = 25.0 - 0 = 25.0 m

But then for (a), it's not zero, which might be surprising, but physically correct.

Perhaps "the tenth floor" means the level of the tenth floor, and the roof is above it.

I think this is correct.

So let's calculate:

m = 61.2 kg
g = 9.8 m/s²

(a) h = 2.5 m
PE = 61.2 * 9.8 * 2.5

First, 61.2 * 2.5 = 153
Then 153 * 9.8 = 153*10 - 153*0.2 = 1530 - 30.6 = 1499.4 J ≈ 1500 J (2 sig figs) or 1.50 × 10^3 J

(b) h = 12.5 m
PE = 61.2 * 9.8 * 12.5

61.2 * 12.5 = 61.2 * 12 + 61.2 * 0.5 = 734.4 + 30.6 = 765
Then 765 * 9.8 = 765*10 - 765*0.2 = 7650 - 153 = 7497 J ≈ 7500 J or 7.50 × 10^3 J

(c) h = 25.0 m
PE = 61.2 * 9.8 * 25.0

61.2 * 25 = 1530
1530 * 9.8 = 1530*10 - 1530*0.2 = 15300 - 306 = 14994 J ≈ 15000 J or 1.50 × 10^4 J

Sig figs: all inputs have 3 sig figs (61.2, 2.50, etc.), g=9.8 has 2, but often in such calculations, we use g=9.80 or accept 3 sig figs.

So we'll report with 3 sig figs.

(a) 1500 J (but 1499.4, so 1.50 × 10^3 J)
(b) 7500 J (7497, so 7.50 × 10^3 J)
(c) 15000 J (14994, so 1.50 × 10^4 J)

But 15000 has 2 sig figs if written as 15000, so better to use scientific notation.

Since the height for (a) is 2.5 m, which has 2 sig figs (2.5, not 2.50? The problem says "2.50 m", so 3 sig figs.

"Each storey is 2.50 m high" — so 3 sig figs.

So h for (a) = 2.50 m (since 25.0 - 22.5 = 2.5, but 22.5 is 9*2.50=22.50, 25.00 - 22.50 = 2.50 m)

Similarly, (b) 25.00 - 12.50 = 12.50 m (sixth floor at 5*2.50=12.50 m)
(c) 25.00 m

So h values are 2.50 m, 12.50 m, 25.00 m

Then PE = mgh

(a) 61.2 * 9.8 * 2.50

Calculate step by step:

61.2 * 2.50 = 153.0
153.0 * 9.8 = 1499.4 J → with sig figs, 9.8 has 2, so 1500 J or 1.5 × 10^3 J

But to be precise, let's use g=9.80 for consistency.

In many problems, they use g=9.8 and expect calculation as is.

I think for this, we'll use g=9.8 and report as:

(a) 1500 J
(b) 7500 J
(c) 15000 J

But let's calculate exactly:

(a) 61.2 * 9.8 * 2.5 = let's do 61.2 * 2.5 = 153, 153 * 9.8 = 1499.4 → 1.50 × 10^3 J (rounded to 3 sig figs)

(b) 61.2 * 9.8 * 12.5 = 61.2 * 12.5 = 765, 765 * 9.8 = 7497 → 7.50 × 10^3 J

(c) 61.2 * 9.8 * 25 = 61.2 * 25 = 1530, 1530 * 9.8 = 14994 → 1.50 × 10^4 J

Yes.

So:

(a) 1.50 × 10³ J
(b) 7.50 × 10³ J
(c) 1.50 × 10⁴ J

✔ Final Answer for Problem 6:
(a) 1500 J
(b) 7500 J
(c) 15000 J

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Problem 7:
> A 10 000 kg airplane lands, descending a vertical distance of 10 km while travelling 100 km measured along the ground. What is the plane’s loss of potential energy?

Loss of potential energy depends only on vertical descent, not horizontal distance.

ΔPE = mgΔh

m = 10,000 kg
g = 9.8 m/s²
Δh = 10 km = 10,000 m (since 1 km = 1000 m)

So ΔPE = 10000 * 9.8 * 10000 = ?

First, 10000 * 10000 = 100,000,000 = 10^8
Then * 9.8 = 9.8 × 10^8 J

Calculate: 10,000 × 10,000 = 100,000,000
100,000,000 × 9.8 = 980,000,000 J = 9.8 × 10^8 J

Sig figs: 10,000 kg (1 or 5? probably 1 or 2; 10 000 could be 1 sig fig if no decimal, but in context, likely 2 or 5. 10 km has 1 or 2 sig figs. 10 km is 1.0 × 10^4 m if 2 sig figs.

Assume 2 sig figs for 10,000 kg and 10 km.

So m = 1.0 × 10^4 kg
Δh = 1.0 × 10^4 m
g = 9.8 m/s² (2 sig figs)

So ΔPE = 1.0e4 * 9.8 * 1.0e4 = 9.8e8 J → with 2 sig figs, 9.8 × 10^8 J is already 2 sig figs.

9.8 has 2, so answer is 9.8 × 10^8 J

✔ Final Answer for Problem 7: 9.8 × 10⁸ J

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Problem 8:
> Calculate the work done by a 2.4 N force pushing a 400 g sandwich across a table 0.75 m wide.

Mass is given, but for work done by the force, we don't need mass, unless there's friction, but here it's just the work done by the applied force.

So W = F × d = 2.4 N × 0.75 m

Calculate: 2.4 × 0.75

2.4 × 0.75 = 2.4 × 3/4 = (2.4/4)*3 = 0.6 * 3 = 1.8 J

Or: 2.4 × 0.75 = 2.4 × (0.7 + 0.05) = 2.4*0.7=1.68, 2.4*0.05=0.12, total 1.80 J

Sig figs: 2.4 (2 sig figs), 0.75 (2 sig figs) → answer 1.8 J (2 sig figs)

The mass 400 g is irrelevant for this calculation, as work done by the force is force times distance, assuming the force is in the direction of motion.

✔ Final Answer for Problem 8: 1.8 J

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Now, compiling all final answers:

1. 47.1 m
2. 12 J
3. (a) 225,000 J (b) 225,000 J
4. 0 J
5. 58,000 J
6. (a) 1500 J (b) 7500 J (c) 15000 J
7. 9.8 × 10⁸ J
8. 1.8 J

For problem 3, I'm still unsure, but I think it's correct as is.

For problem 6, I assumed the heights as per standard interpretation.

Let me write the final answers clearly.

Final Answer:
1. 47.1 m
2. 12 J
3. (a) 225000 J (b) 225000 J
4. 0 J
5. 58000 J
6. (a) 1500 J (b) 7500 J (c) 15000 J
7. 980000000 J or 9.8 \times 10^8 J
8. 1.8 J

To box them as per instruction.

Since the user didn't specify which problem, but the image has multiple, I'll provide all.

But the instruction says "solve the problem", implying one, but there are eight. Probably solve all.

In the response, I'll list them.

For the Final Answer section, since it's multiple, I'll write:

Final Answer:
1. 47.1 m
2. 12 J
3. (a) 225000 J, (b) 225000 J
4. 0 J
5. 58000 J
6. (a) 1500 J, (b) 7500 J, (c) 15000 J
7. 9.8 \times 10^8 J
8. 1.8 J

To make it neat.