WTMA - Wiesen Test of Mechanical Aptitude Springs and Elasticity Questions and Answers — Questions and Answers
Question 1: A spring has a 10 lb weight attached to it, causing it to stretch 2 inches from its natural position. If the 10 lb weight is replaced with a 20 lb weight, how much will the spring stretch from its original, unstretched position, assuming it remains within its elastic limit?
- 8 inches
- 2 inches
- 3 inches
- 4 inches (Correct answer)
Correct answer: 4 inches
According to Hooke's Law, the force (F) required to stretch or compress a spring by some distance (x) is directly proportional to that distance (F = kx). By doubling the force (from 10 lbs to 20 lbs), the displacement (stretch) will also double, from 2 inches to 4 inches.
Question 2: Which of the following statements best describes a spring with a high spring constant (k)?
- It requires a large force to produce a small amount of stretch or compression. (Correct answer)
- It is very long when in its natural, un-stretched state.
- It is very easy to stretch or compress.
- It stores less potential energy than a spring with a low 'k' for the same amount of stretch.
Correct answer: It requires a large force to produce a small amount of stretch or compression.
The spring constant, 'k', is a measure of the stiffness of a spring. A high 'k' value indicates a very stiff spring, which means a large amount of force is needed to deform it (stretch or compress) by even a small distance.
Question 3: A heavy block is placed on a single compression spring, causing it to compress by 4 cm. If an identical second spring is placed next to the first (in parallel), and the same block is placed on top of both springs, what will be the new compression distance?
- 4 cm
- 2 cm (Correct answer)
- 8 cm
- 1 cm
Correct answer: 2 cm
When two identical springs are placed in parallel, they share the load. This doubles the effective spring constant of the system, making it twice as stiff. Since the force (the block's weight) remains the same but the stiffness is doubled, the compression distance will be halved. The new compression will be 4 cm / 2 = 2 cm.
Question 4: What is the term for the point beyond which a spring or other elastic material, when stretched, will not return to its original shape after the deforming force is removed?
- Breaking Point
- Elastic Limit (Correct answer)
- Proportional Limit
- Yield Strength
Correct answer: Elastic Limit
The elastic limit is the maximum stress a material can withstand without undergoing permanent (plastic) deformation. If a spring is stretched beyond this point, it enters the plastic deformation region and will be permanently deformed.
Question 5: Two identical springs are connected end-to-end (in series). A 5 kg mass hung from a single one of these springs causes it to stretch by 10 cm. What will be the TOTAL stretch when the same 5 kg mass is hung from the two-spring series combination?
- 10 cm
- 5 cm
- 20 cm (Correct answer)
- 15 cm
Correct answer: 20 cm
In a series combination, the force (the 5 kg weight) is transmitted through each spring. Since both springs are identical, each will feel the full force and stretch by its characteristic amount, which is 10 cm. The total stretch is the sum of the individual stretches: 10 cm + 10 cm = 20 cm.
Question 6: A spring is compressed by a distance 'x', storing an amount of elastic potential energy 'E'. If the same spring is instead compressed by a distance of '2x', how much elastic potential energy will it store?
- E (the same amount)
- 2E (double the amount)
- 4E (quadruple the amount) (Correct answer)
- E/2 (half the amount)
Correct answer: 4E (quadruple the amount)
The formula for elastic potential energy stored in a spring is PE = ½kx². This shows that the energy is proportional to the square of the displacement (x²). If you double the displacement from 'x' to '2x', the new energy will be proportional to (2x)², which is 4x². Therefore, the stored energy is quadrupled (multiplied by 4).
WTMA - Wiesen Test of Mechanical Aptitude Springs and Elasticity Questions and Answers
A spring has a 10 lb weight attached to it, causing it to stretch 2 inches from its natural position.
If the 10 lb weight is replaced with a 20 lb weight, how much will the spring stretch from its original, unstretched position, assuming it remains within its elastic limit?