Elastic Potential Energy in a Stretched Wire - UNSOLVED PRACTICE SET
Chapter: Mechanical Properties of Solids | Topic: Elastic Potential Energy in a Stretched Wire
ELASTIC POTENTIAL ENERGY IN A STRETCHED WIRE - UNSOLVED PRACTICE SET
Topic: Elastic Potential Energy in a Stretched Wire
Multiple Choice Questions
Q1. The elastic potential energy stored in a stretched wire is:
- ½ × stress × strain × volume
- stress × strain × volume
- ½ × Young's modulus × strain²
- Both (a) and (c) are correct
Q2. The energy stored per unit volume in a stretched wire is:
- ½ × stress × strain
- stress × strain
- ½ × Young's modulus
- ½ × strain²
Q3. For a spring obeying Hooke's law, the potential energy stored when extended by x is:
- ½ kx
- ½ kx²
- kx²
- ½ k²x
Q4. The work done in stretching a wire is stored as:
- Kinetic energy
- Thermal energy
- Elastic potential energy
- Chemical energy
Q5. If a wire is stretched to double its length (within elastic limit), the energy stored becomes:
- Double
- Four times
- Same
- Eight times
Q6. The area under the force-extension graph for a wire represents:
- Young's modulus
- Work done in stretching
- Breaking stress
- Elastic limit
Short Answer Questions
Q7. Derive the expression for elastic potential energy stored in a stretched wire. Show that it equals ½ FΔL.
Q8. A wire of length L and cross-sectional area A is stretched by an amount l. Show that the energy stored per unit volume is ½ × Y × (l/L)².
Q9. A spring with force constant 200 N/m is stretched by 5 cm. Calculate the energy stored in the spring.
Q10. In your school, a student stretches two identical springs — one by 2 cm and another by 4 cm. Compare the energy stored in each. What does this tell you about how energy depends on extension?
Q11. A wire of length 2 m and cross-sectional area 10⁻⁶ m² is stretched by 2 mm. Young's modulus is 2 × 10¹¹ Pa. Calculate the energy stored in the wire.
Q12. Why is the factor ½ present in the expression for elastic potential energy? Explain using the force-extension graph.
Long Answer Questions
Q13. Derive the expression for elastic potential energy stored in a stretched wire starting from first principles. Show that:
(a) U = ½ FΔL
(b) U = ½ × stress × strain × volume
(c) U = ½ Y(strain)² × volume
(d) U = ½ (stress)²/Y × volume
Explain why the energy density is given by ½ σ ε and discuss its significance.
Q14. A steel wire of length 3 m and diameter 1 mm is stretched by a load of 10 kg.
(a) Calculate the extension of the wire. (Y = 2 × 10¹¹ Pa)
(b) Calculate the elastic potential energy stored.
(c) Calculate the energy stored per unit volume.
(d) If this energy were converted entirely to heat, calculate the temperature rise. (Specific heat capacity of steel = 460 J/kg·K, density = 7860 kg/m³)
(e) If the load is suddenly removed, what happens to this energy?
Q15. Compare the energy storage in different systems:
(a) A steel spring compressed by 5 cm
(b) A rubber band stretched by 5 cm
(c) A compressed gas in a cylinder
(d) A raised weight
For each system, discuss:
(i) The formula for stored energy
(ii) Which is most efficient for energy storage
(iii) Practical applications
(iv) Advantages and limitations
Application-Based Problems
Q16. In a school experiment, a student investigates the energy stored in stretched rubber bands of different cross-sections:
Table
Rubber Band Length (cm) Cross-section (mm²) Extension (cm) Force at Extension (N)
A 10 2 2 4
B 10 4 2 8
C 20 2 2 2
(a) Calculate the energy stored in each rubber band.
(b) Calculate the energy stored per unit volume for each.
(c) Which rubber band stores the most energy per unit volume?
(d) If rubber band A is used in a catapult and stretched by 4 cm instead of 2 cm, how much more energy is stored?
(e) Discuss why rubber is preferred over steel for applications like catapults and slingshots.
Q17. A 2 kg mass is attached to a steel wire of length 4 m and cross-sectional area 0.5 mm². The mass is pulled sideways until the wire makes an angle of 30° with the vertical and then released.
(a) Calculate the extension of the wire when the mass is at the lowest point.
(b) Calculate the elastic potential energy stored in the wire at this point.
(c) Calculate the speed of the mass as it passes through the lowest point.
(d) Calculate the maximum angle on the other side if energy is conserved.
(e) Discuss why the actual angle will be slightly less than calculated.
Q18. A bow used in archery can be modelled as a spring. When drawn by 50 cm, it exerts an average force of 200 N.
(a) Calculate the effective force constant of the bow.
(b) Calculate the energy stored in the drawn bow.
(c) If this energy is transferred to an arrow of mass 30 g, calculate the arrow's speed.
(d) If the bow is drawn by 75 cm instead, calculate the new energy and speed.
(e) Discuss why real bows do not obey Hooke's law perfectly and how this affects the calculation.