Question
Download Solution PDFA rod of length 2 m and cross sectional area 50 mm2 is stretched by a load of 100 N. If the change in length of the rod by this load is 0.02 m then find the potential energy density stored in the rod.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFCONCEPT:
Stress:
- Stress is defined as the internal resistance force per unit cross-sectional area of the body.
⇒
Where F = applied load and A = cross-sectional area
Strain:
- When a load is applied to a body it gets deformed and its length changes.
- It is defined as the ratio of the change in length to the original length.
⇒
Elastic potential energy:
- Elastic potential energy is the energy stored as a result of applying a force to deform an elastic object.
- This deformation could involve compressing, stretching, or twisting the object.
- The elastic potential energy stored in the body is given as,
⇒
⇒
- So the potential energy density is given as,
⇒
CALCULATION:
Given l = 2 m, A = 50 mm2 = 50×10-6 m2, F = 100 N and Δl = 0.02 m
- The stress is given as,
⇒
⇒
⇒ σ = 2 × 106 N/m2
- The strain is given as,
⇒
⇒ ϵ = 0.01
- The potential energy density stored in the rod is given as,
⇒
⇒
⇒ u = 104 J/m3
- Hence, option 2 is correct.
Last updated on Jul 4, 2025
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