• Physics 9702 Doubts | Help Page 51

Question 299: [Forces > Hooke’s law]
A wire that obeys Hooke’s law is of length x₁ when it is in equilibrium under a tension T₁; its length becomes x₂ when the tension is increased to T₂. What is the extra energy stored in the wire as a result of this process?
A ¼ (T₂ + T₁) (x₂ – x₁)
B ¼ (T₂ + T₁) (x₂ + x₁)
C ½ (T₂ + T₁) (x₂ – x₁)
D ½ (T₂ + T₁) (x₂ + x₁)
E (T₂ – T₁) (x₂ – x₁)

Reference: Past Exam Paper – J81 / II / 39 & N 87 / I / 30



Solution 299:
Answer: C.
Since the wire obeys Hooke’s law, the graph of force against extension would be a straight line with positive gradient.

Energy stored in the wire is given by the area under the graph. If we want to know the extra energy stored from one extension to another, we need to find the area under the graph between these 2 extensions.

Here, the area under graph between x1 and x2 is given by the area of a parallelogram.
Area of parallelogram = ½ (sum of parallel sides) (height)
Extra energy stored = ½ (T₂ + T₁) (x₂ – x₁)

 







Question 300: [Hooke’s law > Spring constant]
Diagram shows structure of part of a mattress.

Manufacturer wants to design softer mattress (one which will compress more for the same load).
Which change will not have the desired effect?
A using more layers of springs
B using more springs per unit area
C using springs with a smaller spring constant
D using springs made from wire with a smaller Young modulus

Reference: Past Exam Paper – June 2011 Paper 12 Q24



Solution 300:

Answer: B.

We need to know what will NOT have the desired effect of increasing the compression for the same load.

Hooke’s law: F = ke

where F is the load, k the spring constant and e the extension / compression.

For springs in series and in parallel, the following formulae for the ‘effective’ spring constant apply:

Springs in parallel:    k_eff = k₁ + k₂ + ….

Springs in series:       1/k_eff = 1/k₁ + 1/k₂ + ….

Adding more layers implies adding more springs in series while adding more springs in each layer implies adding springs in parallel.

Therefore, it can be seen that (from the formulae) adding springs in parallel increases the effective spring constant while adding springs in series decreases k_eff (for spring of same material).

For choice A, k_eff is decreased and since compression e = F / k, the compression e would increase. So, A is incorrect.

Choice B is correct since k_eff would increase, causing the compression e to decrease.

Choice C is also incorrect since a smaller spring constant k means that the compression e is greater {since e = F / k}.

Young modulus E = stress / strain = (F/A) / (e/L) = FL / Ae. So, compression e = FL / AE. If Young modulus E is smaller, the compression e is bigger. So, choice D is incorrect.

Question 301: [Forces > Hooke’s law]
The following force-extension graphs are drawn to same scale.
Which graph represents the deformed object with greatest amount of elastic potential energy?

Reference: Past Exam Paper – November 2011 Paper 12 Q23



Solution 301:

Answer: B.

The elastic potential energy is represented by the area under the force-extension graph.

The graph is the greatest area is B.

Question 302: [Forces > Hooke’s law]
A door is fitted with a spring-operated latch as shown.

The latch is well-oiled so friction is negligible.
When the latch is pushed in, the spring becomes compressed but remains within its elastic limit.
The latch is then suddenly released.
Which graph best shows how the acceleration a of the latch varies with the distance x it moves before it is stopped?

Reference: Past Exam Paper – N97 / I / 4



Solution 302:
Answer: A.
Since the spring remains within the elastic limit, it obeys Hooke’s law – that is the compression is proportional to the force applied.

From the diagram, when x = 0, the spring is compressed to the maximum. So, the force required for such compression is maximum. From Newton’s law, Force = ma. When the force is maximum, the acceleration is also highest.
Therefore, when x = 0 [compression is maximum], the acceleration a is greatest. [C and D are incorrect]

Hooke’s law: F = ke
Newton’s law: F = ma
[Note that distance x is zero when the compression e is maximum and x is maximum when the compression e is zero. So, x is proportional to (1/e).]

So, ma = ke. That is, the acceleration a and the compression e are linearly proportional. [B is incorrect since the relationship not linear]








Question 303: [Thermodynamics > Ideal gas law]
Suppose that a tank contains 680 m³ of neon at an absolute pressure of 1.01x10⁵ Pa. The temperature is changed from 293.2 to 294.3 K. What is the increase in the internal energy of the neon?

Reference: ???



Solution 303:
Neon is an ideal gas. Its internal energy is associated only with the kinetic energy of the molecules. [Internal energy is the sum of potential and kinetic energy of the molecules in random motion. In an ideal gas, it is assumed that there is no potential energy between the molecules]

Since the neon gas is in a tank (which is rigid) the volume of the gas will not change. So, the work done on the surrounding is zero.

As stated before, the internal energy of an ideal gas depends only on the kinetic energy of the molecules of the gas.
Kinetic energy of 1 molecule of ideal gas = (3/2) kT
Kinetic energy of N molecules of ideal gas = (3/2) NkT
where k is the Boltzmann constant (= 1.38x10^-23 JK^-1)

The change in kinetic energy for a temperature change of ΔT is given by (3/2) NkΔT. This is also the change in internal energy of an ideal gas with N molecules.

The gas is at room temperature and pressure (r.t.p) in the tank (from the values of temperature and pressure given).

At r.t.p, 1 mole of ideal gas occupies a volume of 24dm³
1 mole of an neon contains 6.02x10²³ molecules
1 m³ = 1000 dm³
680 m³ = 680 000 dm³

A volume of 24dm3 contains 6.02x10²³ molecules
Number of neon molecules in tank, N = (680 000/ 24) x 6.02x10²³

Change in temperature ΔT = 294.3 – 293.2 = 1.1 K

Increase in internal energy, ΔU = Increase in kinetic energy of N molecules of neon
Increase in kinetic energy of N molecules of neon = (3/2) NkΔT
ΔU = (3/2) [(680 000/ 24) x 6.02x10²³] (1.38x10^-23) (1.1) = 3.9x10⁵J