Asked by k
Write the answers to each question on a separate sheet of paper.
1. A worker uses a cart to move a load of bricks weighing 680 N a
distance of 10 m across a parking lot. If he pushes the cart with a
constant force of 220 N, what amount of work does he do? Show
your work.
2. A girl lifts a 100-N load a height of 2.0 m in a time of 0.5 s. What
power does the girl produce? Show your work.
3. The input force of a pulley system must move 6.0 m to lift a 3000-
N engine a distance of 0.50 m. What is the IMA of the system?
Show your work.
4. A 16-N force applied to the handle of a door produces a 30-N
output force. What is the AMA of the handle? Show your work.
5. A force of 12 N is applied to the handle of a screwdriver being
used to pry off the lid of a paint can. As the force moves through a
distance 0.3 m, the screwdriver does 32 J of work on the lid. What
is the efficiency of the screwdriver? Show your
All Answers 1
Answered by
GPT-5 mini
AI
1)
Work = force × distance
W = (220 N)(10 m) = 2200 N·m = 2200 J
Answer: 2.20 × 10^3 J
2)
Work = weight × height = (100 N)(2.0 m) = 200 J
Power = work / time = 200 J / 0.50 s = 400 W
Answer: 400 W
3)
IMA = distance input / distance output = 6.0 m / 0.50 m = 12
Answer: IMA = 12
4)
AMA = output force / input force = 30 N / 16 N = 1.875 ≈ 1.88
Answer: AMA ≈ 1.88
5)
Input work = F × d = (12 N)(0.30 m) = 3.6 J
Efficiency = (work out / work in) × 100% = (32 J / 3.6 J) × 100% ≈ 888.9%
Answer: ≈ 8.89 × 10^2 % (≈ 889%)
Note: An efficiency >100% is physically impossible, so the given numbers are inconsistent (likely a mistake in the problem statement—either the 32 J is actually the input work or one of the other values is wrong).
Work = force × distance
W = (220 N)(10 m) = 2200 N·m = 2200 J
Answer: 2.20 × 10^3 J
2)
Work = weight × height = (100 N)(2.0 m) = 200 J
Power = work / time = 200 J / 0.50 s = 400 W
Answer: 400 W
3)
IMA = distance input / distance output = 6.0 m / 0.50 m = 12
Answer: IMA = 12
4)
AMA = output force / input force = 30 N / 16 N = 1.875 ≈ 1.88
Answer: AMA ≈ 1.88
5)
Input work = F × d = (12 N)(0.30 m) = 3.6 J
Efficiency = (work out / work in) × 100% = (32 J / 3.6 J) × 100% ≈ 888.9%
Answer: ≈ 8.89 × 10^2 % (≈ 889%)
Note: An efficiency >100% is physically impossible, so the given numbers are inconsistent (likely a mistake in the problem statement—either the 32 J is actually the input work or one of the other values is wrong).
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