# Elasticity: Problems on Work done and Strain Energy

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#### Example – 1:

• Find the work done in stretching a wire of length 2 m and of sectional area 1 mm² through 1 mm if Young’s modulus of the material of the wire is 2  × 1011 N/m².
• Solution:
• Given: Area  = A = 1 mm² = 1 × 10-6 m², Length of wire = L = 2m, Extension in wire = l = 1mm = 1 × 10-3 m,  Young’s modulus  = Y  =2  × 1011 N/m².
• To Find: Work done = W =?

Young’s modulus of elasticity = Y = FL/Al

∴  F = YAl/L

∴   F = ( 2  × 1011 × 1 × 10-6 × 1 × 10-3 )/2

∴   F = 100 N

Now Work done in stretching wire = ½ Load  ×Extension

∴  Work done = ½ × 100 × 1 × 10-3

∴  Work done = 0.05 J

Ans: Work done in stretching wire is 0.05 J

#### Example – 2:

• Calculate the work done in stretching a wire of length 3 m and cross-sectional area 4 mm² when it is suspended from a rigid support at one end and a load of 8 kg is attached at the free end. Y = 12 × 1010 N/m² and g = 9.8 m/s² .
• Solution:
• Given: Area  = A = 4 mm² = 4 × 10-6 m², Length of wire = L = 3m, Load = 8 kg-wt = 8 × 9.8 N,  Young’s modulus  = Y  = 12 × 1010 N/m².
• To Find: Work done = W =?

Young’s modulus of elasticity = Y = FL/Al

∴  l = FL/AY

∴  l = (8 × 9.8 × 3)/( 4 × 10-6 × 12 × 1010)

∴  l = 4.9 × 10-4 m

Now Work done in stretching wire = ½ Load  ×Extension

∴  Work done = ½ × 8 × 9.8 × 4.9 × 10-4

∴  Work done = 1.921 × 10-2 J = 0.0192 J

Ans: Work done in stretching wire is 0.0192 J

#### Example – 3:

• When the load on a wire is increased slowly from 3 to 5 kg wt, the elongation increases from 0.6 to 1 mm. How much work is done during the extension? g = 9.8 m/s² .
• Solution:
• Given: Initial Load = F1 = 3 kg wt = 3 × 9.8 N, Final load =F2 =  5 kg-wt = 5 × 9.8 N, Initial extension l1 = 0. 6 mm = 0.6  × 10-3  m = 6  × 10-4  m, Final extension = l2 = 1mm = 1  × 10-3  m = 10  × 10-4  m, g = 9.8 m/s² .
• To Find: Work done = W =?

Work done = W = W2 – W1

∴  Work done = ½ × F2× l2  –  ½ × F1 × l1

∴  Work done = ½ ×(F2 × l2  –   F1 × l1)

∴  Work done = ½ ×(5 × 9.8 × 10  × 10-4  –  3 × 9.8 × 6  × 10-4)

∴  Work done = ½ × 9.8  × 10-4(50  –  18)

∴  Work done = ½ × 9.8  × 10-4  × 32

∴  Work done =1.568  × 10-2  = 0.01568 J

Ans : Work done is  0.01568 J

#### Example – 4:

• A spring is compressed by 1 cm by a force of 3.92 N. What force is required to compress it by 5 cm? What is the work done in this case? Assume the Hooke’s Law.
• Solution:
• Given: Initial Load = F1 = 3.92 N, Initial extension l1 = 1 cm = 1  × 10-2  m , Final extension = l2 = 5 cm = 5  × 10-2  m.
• To Find: Final Load = F2 = ?, Work done = W =?

We have Force constant = K = F/l

Hence  F1/l1 =  F2/l2

Hence  F2   = F× l2  / l1 =  (3.92 × 5  × 10-2)  /(1  × 10-2)

∴  F2   =  (3.92 × 5  × 10-2)  /(1  × 10-2)

Ans : (9.8 N; 0.49)

#### Example – 5:

• A wire 4m long and 0.3 mm in diameter is stretched by a load of 0.8 kg. If the extension caused in the wire is 1.5 mm, find the strain energy per unit volume of the wire.g = 9.8 m/s²
• Solution:
• Given: Length of wire = L = 4m, Diameter = 0.3 mm, Radius of wire = r = 0.3/2 = 0.15 mm = 015 × 10-3 m = 1.5 × 10-4 m,Area  = Load applied = F = 0.8 kg-wt = 0.8 × 9.8 N, Extension in wire = l = 1.5 mm = 1.5 × 10-3 m, .g = 9.8 m/s².
• To Find: Strain energy per unit volume = dU/V =?

Strain energy per unit volume =½ × Stress × Strain

∴  dU/V =½ × (F/A) × (l/L)

∴  dU/V =½ × (Fl/AL)

∴  dU/V =½ × (Fl/πr²L)

∴  dU/V =½ × ( 0.8 × 9.8  × 1.5 × 10-3/(3.142 × (1.5 × 10-4)² × 4)

∴  dU/V =½ × ( 0.8 × 9.8  × 1.5 × 10-3/(3.142 × 2.25 × 10-8 × 4)

∴  dU/V = 2.08 × 104   J/m³

Ans :The strain energy per unit volume of the wire  2.08 × 104   J/m³

#### Example – 6:

• Find the energy stored in a stretched brass wire of 1 mm² cross-section and of an unstretched length 1 m when loaded by 2 kg wt. What happens to this energy when the load is removed. Y = 1011 N/m².
• Solution:
• Given: Area  = A = 1 mm² = 1 × 10-6 m², Length of wire = L = 1 m, Load = 2 kg-wt = 2 × 9.8 N,  Young’s modulus  = Y  = 1011  N/m².
• To Find: Energy stored = dU =?

Young’s modulus of elasticity = Y = FL/Al

∴  l = FL/AY

∴  l = (2 × 9.8 × 1)/( 1 × 10-6  × 1011)

∴  l = 1.96  × 10-4 m

Now Work done in stretching wire = ½ Load  ×Extension

∴  Work done = ½ × 2 × 9.8 × 1.96 × 10-4

∴  Work done = 1.921 × 10-3 J

Now energy stored = Work done in stretching wire

Ans:  Energy stored is 1.921 × 10-3 J

#### Example – 7:

• A metal wire of length 2.5 m and are of cross section 1.5 × 10-6 m² is stretched through 2 mm. Calculate work done during stretching. Y = 1.25 × 1011 N/m².
• Solution:
• Given: Area  = A = 1.5 × 10-6 m², Length of wire = L = 2.5 m, Extension = l = 2mm = 2 × 10-3 m, Young’s modulus  = Y  = 1.25 × 1011 N/m².
• To Find: Energy stored = dU =?

Young’s modulus of elasticity = Y = FL/Al

∴  F = YAl/L

∴  F = (1.25 × 1011 × 1.5 × 10-6 × 2 × 10-3 )/2.5

∴  F = 150 N

Now Work done in stretching wire = ½ Load  ×Extension

∴  Work done = ½ × 150 × 2 × 10-3

∴  Work done = 0.150 J

Now energy stored = Work done in stretching wire

Ans:  Energy stored is 0.150 J

#### Example – 8:

• A copper wire is stretched by 0.5% of its length. Calculate the energy stored per unit volume in the wire. Y  = 1.2 × 1011 N/m².
• Solution:
• Given: Strain = l/L  = 0.5 %  = 0.5 × 10-2 = 5 × 10-3, Young’s modulus  = Y  = 1.2 × 1011 N/m².
• To Find: Strain energy per unit volume = dU/V =?

Strain energy per unit volume = dU/V = ½ × (Strain)² × Y

∴   dU/V = ½ × (5 × 10-3)² × 1.2 × 1011

∴   dU/V = ½ × 25 × 10-6 × 1.2 × 1011

∴   dU/V = 1.5 × 106    J/m³

Ans: The strain energy per unit volume of the wire  1.5 × 106    J/m³

#### Example – 9:

• Calculate the strain energy per unit volume in a brass wire of length 2.0 m and cross-sectional area 0.5 mm2, when it is stretched by 2mm and a force of 5 kg-wt is applied to its free end.
• Solution:
• Given: Area  = A = 0.5 mm² = 0.5 × 10-6 m² = 5 × 10-7 m², Length of wire = L = 2.0 m, Extension in wire = l = 2 mm = 2 × 10-3 m,  Load applied = F = 5 kg-wt = 5 × 9.8 N
• To Find: Strain energy per unit volume = dU/V =?

Strain energy per unit volume = dU/V = ½ × Stress × Strain

∴    Strain energy per unit volume = ½ × (F/A) × (l/L)

∴    Strain energy per unit volume = ½ × (Fl/AL)

∴    Strain energy per unit volume = ½ × (5 × 9.8 ×  2 × 10-3)/( 5 × 10-7 × 2)

∴    Strain energy per unit volume = 4.9 × 104  J/m³

Ans: The strain energy per unit volume of the wire  4.9 × 104  J/m³

#### Example – 10:

• Calculate the work done in stretching a wire of length 2 m and cross-sectional area 0.0225 mm² when a load of 100 N is applied slowly to its free end. Young’s modulus of elasticity = 20 × 1010 N/m².
• Solution:
• Given: Area  = A =0.0225 mm² =0.0225 × 10-6 m² = 2.25 × 10-8 m², Length of wire = L = 2 m, Load applied = F = 100 N, Young’s modulus of elasticity = Y = 20 × 1010 N/m².
• To Find: Work done = W =?

Young’s modulus of elasticity = Y = FL/Al

∴  l = FL/AY

∴  l = (100 × 2)/( 2.25 × 10-8 × 20 × 1010)

∴  l = 4.444 × 10-2 m

Now Work done in stretching wire = ½ Load  ×Extension

∴  Work done = ½ × 100 × 4.444 × 10-2

∴  Work done = 2.222 J

Ans: Work done in stretching wire is 2.222 J

#### Example – 11:

• A uniform steel wire of length 3 m and area of cross-section 2 mm² is extended through 3mm. Calculate the energy stored in the wire, if the elastic limit is not exceeded. Young’s modulus of elasticity = Y = 20 × 1010
• Solution:
• Given: Area  = A =2 mm² =2 × 10-6 m² , Length of wire = L = 3 m, Extension = l = 3 mm = 3 × 10-3 m, Young’s modulus of elasticity = Y = 20 × 1010 N/m².
• To Find: Energy stored  = dU =?

Young’s modulus of elasticity = Y = FL/Al

∴  F = YAl/L

∴   F = ( 20  × 1010 × 2 × 10-6 × 3 × 10-3 )/3

∴   F = 400 N

Now Work done in stretching wire = ½ Load  ×Extension

∴  Work done = ½ × 400 × 3 × 10-3

∴  Work done = 0.6 J

Energy stored = work done in stretching wire = 0.6 J

Ans: Energy stored is 0.6 J

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