 # JAMB 2022 PHYSICS PAST QUESTION 22 ELASTIC LIMIT AND HOOKE’S LAW

## ELASTIC PROPERTIES OF SOLIDS

Young Modulus
Young Modulus, y is defined as the ratio of tensile stress to tensile strain.
Y = tensile stress/tensile strain
Tensile strain is the ratio of the extension of a material to the cross sectional area of the material.
Strain = change in extension/ original length = e/L
Tensile strain has no unit
Tensile stress is the ratio of the force acting on a material to the cross sectional area of the material
Stress= force/area
Therefore, Y = tensile stress/tensile strain
= Force/area ÷ extension/original length
F/A × L/e = FL/Are
The S.I unit for y is N/m²
WORK DONE IN SPRINGS AND ELASTIC STRINGS

If an applied force, F caused an elastic spring of original length, l, to undergo an extension or comprehension, e, then the average force is;
= (0+F)/2 = ½F
Work done = force × distance
= ½F × e
W = ½Fe or ½Ke²
This is the work done by a spring when compressed or extended by a force. The energy stored in a spring or string is also given by W = ½Fe or ½Ke². The unit of work or energy is joules, j.

1. A spiral spring balance is 25.0cm long when 5N hangs on it and 30.0cm when the weight is 10N. What is the length of the spring if the weight is 2N, assuming Hooke’s law is obeyed?

Solution
Let the original length of the spiral spring be L
Extension, e = New length resulting from applied force – original length
First case: extension, e = 25-L
Applied force, F = 5N
2nd case: extension, e = 30-L
Applied force, F = 10N
Substitute into F = ke for 1st and 2nd cases.
1st. 5 = (25-L)k……………..(1)
2nd. 10 = (30-L)k…………..(2)
Make the subject of the formula,
K = 5/(25-L)………..(3)
K = 10/(30-L)………..(4)
Equate (3) and (4)
5/(25-L) = 10/(30-L)
Cross multiply both sides together
150-5L = 250-10L
Collect like terms
5L = 100
L = 20cm
Extension, e = 25.0 – L
25.0 – 20 = 5.0cm
= 0.05m( convert to meter)
Recall that, F = ke ( applied force is 5N)
K = f/e; = 5/0.05
K = 100N/m
What is the length of the spring if the weight is 2N,
e = F/k;
2/100 = 0.02m
= 2cm
New length of spring = Original length (L) + Extension (e)
20.0 + 2 = 22.0cm

1. A catapult used to hold a stone of mass 500g is extended by 20cm with an applied force F. Of the stone leaves with a velocity of 40ms–¹, the value of F is
A. 4.0×10⁴N. B. 4.0×10³N. C. 2.0×10³N. D. 4.0×10²N
Solution
Mass = 500g = 0.5Kg( convert to kilogram) , e = 20cm = 0.2m, velocity= 40ms–¹
F = ke
Work done = ½mv²
½×0.5×40² = ½×0.5×1600
W.D = 400j
Recall that, W.D = ½Fe
400 = ½×F×0.2
F = 400/0.1
F = 4,000N or 4×10³N
2. On top of a spiral spring of force constant 500Nm–¹ is placed a mass of 5×10–³Kg. If the spring is compressed downwards by a length of 0.02m and then released, calculate the height to which the mass is projected. (g=10ms–²)
A. 8m B. 4m C. 2m D. 1m
Solution
Force constant, k= 500Nm–¹,. Mass = 5×10–³kg, length = 0.02m, height = ?
The work done in compressing the spring is equal to the potential energy stored in the spring.
½ke² = mgh
½×500×(0.02)² = 0.005×10×h
0.1 = 0.05h
H = 0.1/0.05
= 2m
3. A load of 5N gives an extension of 0.56cm in a wire which obeys Hooke’s law. What is the extension caused by a load of 20N?
1.12cm B. 2.41cm C. 2.24cm D. 2.52cm
Solution
Force = 5N, e = 0.56cm( no need for conversion, since its unit of answer are in cm),
First let’s find the constant value
K = F/e = 5/0.56 = 8.93N/cm
e = F/K = 20/8.93 = 2.24cm
4. Use the following data to determine the length, L of a wire when a force of 30N is applied, assuming Hooke’s law is obeyed.

SOLUTION
Force applied = 5-0 = 5N
Extension Length of wire (mm) = 500.5-500.0= 0.5mm
Constant, K = F/e = 5/0.5 = 10M/m
extension = F/k = 30/10 = 3mm
L = original length + Extension
500.0+3 = 503.0mm

## EXERCISE TWO(2)

1. If a force of 50N stretches a wire from 20m to 20.01m, what is the amount of force required to stretch the same material from 20m to 20.05m
A. 100N. B. 50N. C. 250N. D. 200N
Solution
Force = 50N, extension 1 = 20.01-20 = 0.01m,
Find the constant k
K = F/e
50/0.01 = 5000N/m
what is the amount of force required to stretch the same material from 20m to 20.05m
Extension 2 = 20.05-20.0 = 0.05m
F = Ke
5000×0.05 = 250N. (C)
2. The extension in a spring when 5g weight was hung from it was 0.56cm. If Hooke’s law is obeyed, what is the extension caused by a load of 20g weight?
A. 1.12cm B. 2.24cm. C. 2.52cm. D. 2.80cm
Solution
Mass = 5g = 0.005kg, extension = 0.56cm
Convert mass to newton
Force= 0.005×10; = 0.05N
Find constant, k = F/e
0.05/0.56 = 0.089N/m
what is the extension caused by a load of 20g weight?
20g = 0.02×10
0.2N
extension,e = F/k
0.2/0.089 = 2.24cm
3. A 10g mass placed on the plan of a spring balance cause an extension of 5cm. If a 15g mass is placed on the plan of the same spring balance the extension is?
A. 3.3cm B. 6.5cm. C. 7.5cm. D. 10.8cm. E. 15.0cm
Solution
Do the same thing here, covert gram to newton
Mass = 10g = 0.01kg
Force = 0.01×10 = 0.1N
extension 1,e = 5cm
Find constant k,
K = F/e
0.1/5 = 0.02N/m If a 15g mass is placed on the plan of the same spring balance the extension is?
Force, F = 0.015×10 = 0.15N
Extension, e = F/k
0.15/0.02 = 7.5cm
4. The total length of a spring when a mass of 20g is hung from its end is 14cm, while its total length is 16cm when a mass of 30g is hung from the same end. Calculate the unstretched length of the spring assuming Hooke’s law is obeyed.
A. 9.33cm. B. 10.00cm. C. 10.66cm. D. 12.00cm. E. 15.00cm
Solution
Mass = 20g, force = 0.2N, extension = 14cm
Mass= 30g, force = 0.3N, extension = 16cm
F = Ke
0.2 =k(14-L)……………………………….(1)
0.3 = k(16-L)……………………………….(2)
Make k the subject of the formula
K = 0.2/(14-L)……………………………..(3)
K = 0.3(16-L)………………………………(4)
Equate (3) and (4)
0.2/(14-L) = 0.3(16-L)
3.2-0.2L = 4.2-0.3L
0.3L-0.2L = 4.2-3.2
0.1L = 1.0
L = 1.0/0.1 = 10.00cm
5. A force of 100N stretches an elastic string to a total length of 20cm. If an additional force of 100N stretches the string 5cm further, find the natural length of the spring.
A. 15cm. B. 12cm. C. 8cm. D. 5cm
Solution
F = 100N, original length = 20cm, extension = 5cm
Length = original length- extension
20-5 = 15cm
6. The spiral spring of a spring balance is 25.0cm long when 5N hangs on it and 30.0cm long when the weight is 10N. What is the length of the spring if the weight is 3N assuming Hooke’s law is obeyed?
A. 15.0cm. B. 17.0cm. C. 20.0cm. D. 23.0cm
Solution

Force = 5N, extension = 20.0cm
Force = 10N, extension = 30.0cm
F = Ke
5 =k(20-L)……………………………….(1)
10 = k(30-L)……………………………….(2)
Make k the subject of the formula
K = 5/(25-L)……………………………..(3)
K = 10(30-L)………………………………(4)
Equate (3) and (4)
5/(25-L) = 10(30-L)
150-5L = 250-10L
10L-5L = 250-150
5L = 100
L = 100/5 = 20.00cm
Considering the first case;
Extension, e = 25.0-L = 25-20
5cm = 0.05m
Applied Force = 5N
From, f = Ke, force constant, k = F/e = 5/0.05
= 100N/m
for F = 3N and K = 100N/m
Extension,e = F/k
3/100 = 0.03m = 3cm
Therefore, New length of spring = Original length (L) + Extension (e)
20cm+3cm = 23cm

1. A force of 15N stretches a spring to a total length of 30cm. An additional force of 10N stretches the spring 5cm further. Find the natural length of the spring.
A. 25.0cm. B. 22.5cm. C. 15.0cm. D. 20.0cm
Solution
Force = 15N, extension = 30.0cm
Force = 10N, extension = 5.0cm
Find the natural length of the spring,
Assuming, the both force are 10N then the natural length will be 30-5 = 25cm
So because of the 15N, it should be 22.5cm
2. A piece of rubber 10cm long stretches 6mm when a load of 100N is hung from it. What is the strain?
A. 60. B. 6. C. 6×10–³. D. 6×10–³
Solution
Length, L = 10cm = 100mm
Extension, e = 6mm
Strain = extension/length
6/100 = 0.06 = 6×10–²
3. A load of 20N on a wire of cross-sectional area 8×10–⁷m², produces an extension of 10–⁴m. Calculate Young’s Modulus for the material of the wire if it’s length is 3m.
A. 7.0 × 10¹¹Nm–². B. 7.5 × 10¹¹Nm–². C. 8.5 × 10¹¹Nm–². D. 9.0 × 10¹¹Nm–²
Solution
Force F = 20N, length = 3m
Area = 8×10–⁷m², extension = 10–⁴m.
Young Modulus = FL/Ae
(20×3) ÷ (8×10–⁷ × 10–⁴)
60/8×10–¹¹ = 7.5 × 10¹¹Nm–².
4. The tendon in a man’s leg is 0.01m long. If a force of 5N stretches the tendon by 2.0 × 10–⁵m, calculate the strain on the muscle.
A. 5 × 10⁶. B. 5 × 10². C. 2 × 10–³. D. 2 × 10–⁷
Solution
Strain = extension/length
0.00002/0.01 = 2 × 20–³
5. If the stress on a wire is 10⁷Nm–² and the wire is stretched from its original length of 10.0cm to 10.05cm. The Young modulus of the wire is
A. 5.0×10⁴Nm–². B. 5.0×10⁵Nm–². C. 2.0×10⁸Nm–². D. 2.0×10⁹Nm–²
Solution
Young modulus = stress/strain
Stress = 10⁷Nm–²
Length = 10.0cm
Extension e = 10.05-10.0 = 0.05cm = ,0.0005m
Strain = extension/length
0.0005/0.1 = 0.005
Young modulus = 10⁷/0.005
= 2.0×10⁹Nm–²
6. A spring of force constant 1500Nm–¹ is acted upon by a constant force of 75N. Calculate the potential energy stored in the spring.
A. 1.9j. B. 3.2j. C. 3.8j. D. 5.0j
Solution
Constant, k = 1500Nm–¹. Force F = 75N
Potential energy = ½Fe
Find extension
F = Ke
e = F/K
75/1500 = 0.05m
Potential energy = ½Fe
½×75×0.05 = 1.875J
Approximately = 1.9J
7. A spring of length 25cm is extended to 30cm by a load of 150N attached to one of its ends. What is the energy stored in the spring?
A. 3750J. B. 2500J. C. 3.75J. D. 2.50J
Solution
What is the energy stored in the spring?
Extension = 30-25 = 5cm = 0.05
Energy stored = ½Fe
½×150×0.05 = 3.75J
8. The energy contained in a wire when it is extended by 0.02m by a force of 500N is
A. 5J. B. 10J. C. 10³J D. 10⁴J
Solution
What is the energy stored in the spring?
Extension = 0.02m
Energy stored = ½Fe
½×500×0.02= 5J
9. 50.0kg block is dropped on a spring from a point 10m above ( above figures). If the force constant of the spring is 4.0×10⁴Nm–¹, find the maximum compression of the spring [g = 10m/s²].
A. 1.25m. B. 0.50m. C. 0.25m. D. 0.05m
Solution
Mass = 50.0kg,. Length = 10m, force constant k = 4.0×10⁴Nm–¹, gravity = 10ms–²
½Ke² = mgh
½×40000×e² = 50×10×10
e² = 5000/20000
e = √0.25
e = 0.50m
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### Oluwamuyide Peter

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