# Physics 9702 Doubts | Help Page 51

__Question 299: [Forces > Hooke’s law]__A wire that obeys Hooke’s law is of length x

_{1}when it is in equilibrium under a tension T

_{1}; its length becomes x

_{2}when the tension is increased to T

_{2}. What is the extra energy stored in the wire as a result of this process?

A ¼ (T

_{2}+ T

_{1}) (x

_{2}– x

_{1})

B ¼ (T

_{2}+ T

_{1}) (x

_{2}+ x

_{1})

C ½ (T

_{2}+ T

_{1}) (x

_{2}– x

_{1})

D ½ (T

_{2}+ T

_{1}) (x

_{2}+ x

_{1})

E (T

_{2}– T

_{1}) (x

_{2}– x

_{1})

**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 (T

_{2}+ T

_{1}) (x

_{2}– x

_{1})

__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}= k_{1}+ k_{2}+ ….**Springs in series: 1/k**

_{eff}= 1/k_{1}+ 1/k_{2}+ ….
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

**proportional. [B is incorrect since the relationship not linear]**

__linearly__

__Question 303: [Thermodynamics > Ideal gas law]__Suppose that a tank contains 680 m

^{3}of neon at an absolute pressure of 1.01x10

^{5}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 24dm

^{3}

1 mole of an neon contains 6.02x10

^{23}molecules

1 m

^{3}= 1000 dm

^{3}

680 m

^{3}= 680 000 dm

^{3}

A volume of 24dm3 contains 6.02x10

^{23}molecules

Number of neon molecules in tank, N = (680 000/ 24) x 6.02x10

^{23}

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 6.02x10

^{23}] (1.38x10

^{-23}) (1.1) = 3.9x10

^{5}J

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