Question

45. Consider the situation shown in figure (8-E8). Initially
the spring is unstretched when the system is released
from rest. Assuming no friction in the pulley, find the
maximum elongation of the spring.
GL
m
0000000
Figure 8-E8​

Answer 1

Answer:Let mass of the body be m,

Let mass of the body be m,Assume that the highest elongation in the spring be x .

Let mass of the body be m,Assume that the highest elongation in the spring be x .As given No friction in the pulley , Then for the block , Reduction in the potential energy is equal to potential energy stored in spring .

Let mass of the body be m,Assume that the highest elongation in the spring be x .As given No friction in the pulley , Then for the block , Reduction in the potential energy is equal to potential energy stored in spring .Now, Potential energy in the spring = 1/2 (kx²)

Let mass of the body be m,Assume that the highest elongation in the spring be x .As given No friction in the pulley , Then for the block , Reduction in the potential energy is equal to potential energy stored in spring .Now, Potential energy in the spring = 1/2 (kx²)and change in potential energy = mgx

Let mass of the body be m,Assume that the highest elongation in the spring be x .As given No friction in the pulley , Then for the block , Reduction in the potential energy is equal to potential energy stored in spring .Now, Potential energy in the spring = 1/2 (kx²)and change in potential energy = mgx Hence on equating these two values.

Let mass of the body be m,Assume that the highest elongation in the spring be x .As given No friction in the pulley , Then for the block , Reduction in the potential energy is equal to potential energy stored in spring .Now, Potential energy in the spring = 1/2 (kx²)and change in potential energy = mgx Hence on equating these two values.1/2 (kx²) = mgx

Let mass of the body be m,Assume that the highest elongation in the spring be x .As given No friction in the pulley , Then for the block , Reduction in the potential energy is equal to potential energy stored in spring .Now, Potential energy in the spring = 1/2 (kx²)and change in potential energy = mgx Hence on equating these two values.1/2 (kx²) = mgx .

Let mass of the body be m,Assume that the highest elongation in the spring be x .As given No friction in the pulley , Then for the block , Reduction in the potential energy is equal to potential energy stored in spring .Now, Potential energy in the spring = 1/2 (kx²)and change in potential energy = mgx Hence on equating these two values.1/2 (kx²) = mgx .Hence the maximum elongation in the spring is x = 2mg /k .

Let mass of the body be m,Assume that the highest elongation in the spring be x .As given No friction in the pulley , Then for the block , Reduction in the potential energy is equal to potential energy stored in spring .Now, Potential energy in the spring = 1/2 (kx²)and change in potential energy = mgx Hence on equating these two values.1/2 (kx²) = mgx .Hence the maximum elongation in the spring is x = 2mg /k .Hope it Helps.

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Answer 2

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