Question

Practice Exercises
Exercise 1: A ski lift carries people along a 220-meter cable up the
side of a mountain. Riders are lifted a total of
110 meters in elevation. What is the ideal mechanical
advantage of the ski lift?
Exercise 2: The ideal mechanical advantage of a machine is 3.2 and
its output distance is 2.5 meters. What is the input
distance?
Work output
Work input
8 J
10 J
Work output
Work input
Physical Science Math Skills and Problem Solving Workbook 57



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Exercise 3: Paco rode his skateboard up a ramp for a distance of
3.8 meters. If the ideal mechanical advantage of the
ramp is 2, how far above the ground was Paco?
Exercise 4: The inventor of a new machine claims that its actual
mechanical advantage is 4. Literature on the machine
reports its input distance is 81 meters and its output
distance is 27 meters. Find the machine’s ideal
mechanical advantage, and determine whether the
developer’s claim could be true.
Exercise 5: What is the efficiency of a machine that has work input
of 40 joules and work output of 35 joules?
Exercise 6: Work output of a large machine in a factory is
89,000 joules, and its input is 102,000 joules. Work
output of a similar machine is 92,000 joules, and its
work input is 104,000 joules. Which machine has greater
efficiency?
Exercise 7: Lucas has been challenged to increase the efficiency of a
device he is redesigning. His goal is 90 percent
efficiency. The work input of the device is fixed at
10 joules, so Lucas is trying to improve work output.
How many joules of work output should Lucas aim for
to attain his goal of 90 percent efficiency?

Answer 1

Here are the answers with brief work shown:

1. IMA = input distance / output distance = 220 m / 110 m = 2.

2. Input distance = IMA × output distance = 3.2 × 2.5 m = 8.0 m.

3. Output (height) = input distance / IMA = 3.8 m / 2 = 1.9 m.

4. IMA = 81 m / 27 m = 3. The claimed actual mechanical advantage of 4 is not possible because actual MA cannot exceed ideal MA (friction and losses make AMA ≤ IMA). So the claim is false.

5. Efficiency = (work output / work input) × 100% = (35 J / 40 J) × 100% = 87.5%.

6. Machine A: 89,000 / 102,000 = 0.8725 → 87.3% (approx).
Machine B: 92,000 / 104,000 = 0.8846 → 88.5% (approx).
Machine B is more efficient.

7. Required work output = efficiency × work input = 0.90 × 10 J = 9 J.