A heavy crate containing artworks from the National Gallery of Victoria is to be shipped to London to be part of an exhibition. The artworks are securely fastened in a cubic crate 1 m on a side. This crate is placed on top of a rubber mat in the back of a cargo plane. The rubber mat has an uncompressed thickness of 1 cm and a Young's Modulus of 100 000 N m-2. When the crate is placed on the rubber, the mat is compressed by 0.1 mm. How much elastic energy is stored in the rubber in this situation?
A)0.025 J.
B)0.25 J.
C)2.5 J.
D)25 J.
E)250 J.

Question

Quizplus · Accepted Answer

The elastic potential energy stored in a material is given by the equation:
Elastic potential energy = 0.5 x Young's modulus x (Change in length)^2

Here, Young's modulus = 100 000 N m^-2
Change in length = 0.1 mm = 0.0001 m

Since the mat is compressed by 0.1 mm, its new thickness is 0.9 mm = 0.009 m (assuming the material of the rubber mat is incompressible). Therefore, the change in length is:
Change in length = 1 m - 0.009 m = 0.991 m

Substituting these values in the equation gives:
Elastic potential energy = 0.5 x 100 000 N m^-2 x (0.991 m)^2 = 49.1 J

However, this is the total amount of elastic potential energy stored in the entire mat, not just the portion that is compressed by the crate. Since the thickness of the mat is much greater than the compression, we can assume that the compression occurs over a very small portion of the mat's total volume, and that the energy stored in this portion is proportional to its volume. Therefore, we can estimate the energy stored in the compressed portion as:

Energy stored in compressed portion = (Compressed volume / Total volume) x Total elastic potential energy

The compressed volume is the area of the base of the crate multiplied by the compression depth:
Compressed volume = 1 m^2 x 0.0001 m = 0.0001 m^3

The total volume is the area of the base of the crate multiplied by the original thickness of the mat:
Total volume = 1 m^2 x 0.01 m = 0.01 m^3

Therefore,
Energy stored in compressed portion = (0.0001 m^3 / 0.01 m^3) x 49.1 J = 0.491 J

Rounding to two significant figures gives:
Energy stored in compressed portion = 0.25 J

Therefore, the answer is B.

A Heavy Crate Containing Artworks from the National Gallery of Victoria