Elasticity problems 

The following elasticity problems will help you understand elasticity and Hooke's law.

Problem #1: A spring stretches 5 cm when a load of 20 N is hung on it. If instead, we put a load of 30 N, how much will the spring stretch? What is the spring constant?

Solution: There are a couple of ways to solve this problem.

Way #1: Notice that 30 N = 20 N  + 10 N

20 N creates a stretch of 5 cm. Since 10 N is half of 20 N, then 10 N will create a stretch that is half of 5 cm or 2.5 cm.

Total stretch = 5 cm  + 2.5 cm = 7.5 cm

Way #2: Set up a proportion.

5 cm is to 20 N as 'new stretch' is to 30 N.

As a proportion, we get  
5 / 20
 =
new stretch / 30
As a proportion, we get

5 / 20
 =
new stretch / 30


Cross multiply

5 × 30 = new stretch × 20

150 = new stretch × 20

new stretch =  
150 / 20
 = 7.5 cm

To get the spring constant, make a couple of good observation.

20 = 4 × 5

30 = 4 × 7.5

F = 4 × x

F is the force applied and x is the stretch

The spring constant is k = 4

Problem #2: With a weight of 25 kg, a spring stretches 6 cm. Its elastic limit is reached with a weight of 150 kg. How far did the spring stretch?

Solution

Since 150 kg divided by 25 kg = 6 kg, 150 kg is 6 times bigger.

The stretch will then be 6 times bigger than 6 cm or 36 cm.

Problem #3: A spring has a spring constant that is equal to 3.5. What force (in kilograms) will make it stretch 4 cm? 

F = k × x

F = 3.5 × 4

F = 14 kg

Tough elasticity problems

Problem #4: When the weight hung on a spring is increased by 60 N, the new stretch is 15 cm more. If the original stretch is 5 cm, what is the original weight?

We will need some algebra and a proportion to solve this tough word problem.

Let x be the original weight, then x + 60 is the new weight

If the original stretch is 5 cm, then the new stretch is 20 cm.

x is to x + 60 as 5 cm is to 20 cm.

As a proportion, we get  
x / x + 60
 =
5 / 20

As a proportion, we get

x / x + 60
 =
5 / 20


Cross multiply

x × 20 = 5 × x + 5 × 60

20x = 5x + 300

15x = 300

Since 15 × 20 = 300, the original weight is 20 N

Problem #5: The elastic limit of a spring is reached with a weight of 90 kg. In this situation, the final stretch is 20 more the original. If the original weight is 75 less the final weight, what is the final stretch?

Let x be the original stretch, then x + 20 is the final stretch.

The original weight must be 15 kg if the final weight is 90 kg.

x is to 15 kg as x + 20  is to 90 kg.

x / 15
 = 
x + 20 / 90

Cross multiply

x × 90 = 15 × x + 15 × 20

90x = 15x + 300

75x = 300

Since 75 × 4 = 300, x = 4

The final stretch is then 4 + 20 or 24 cm

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