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Homework Statement
A body of mass (m) is suspended from a spring with spring constant k in configuration (a) and the spring is stretched 0.1m. If two identical bodies of mass (m/2) are suspended from a spring with the same spring constant k in configuration (b), how much will the spring stretch? Explain your answer.
Homework Equations
F = -kx (Hooke's Law)
The Attempt at a Solution
Since in the first case the mass is at rest,
by mg - k(0.1) = 0, i calculated that k = 10mg
In the second case, since the bodies suspended have a mass of (m[tex]\div[/tex]2), their weight is halved too. Therefore the spring will stretch for 0.1[tex]\div[/tex]2 = 0.05m on each side.
i.e. It stretches for 0.1m (0.05m each on left and right)
However, according to the solution, the answer is 0.05m.
It is also stated that "if you have problems, substitute one of the two masses with a fixed hook; does the spring change its length?"
What difference does that make?
Attached is the diagram describing the setup