A block that has a mass equal to m1 is supported from below by a frictionless ho

Question

A block that has a mass equal to m1 is supported from below by a frictionless horizontal surface. The block, which is attached to the end of the horizontal spring that has a force constant k, oscillates with an amplitude A. When the spring is at its greatest extension and the block is instantaneously at rest, a second block of mass m2 is placed on top of it.
a) what is the smallest value for the coefficient of static friction mue(s) such that the second object does not slip on the first.
b) explain how the total mechanical energy E, the amplitude A, the angular frequency omega, and the period T of the system are affected by the placing of m2 on m1, assuming that the coefficient of friction is great enough to prevent slippage A block that has a mass equal to m1 is supported from below by a frictionless horizontal surface. The block, which is attached to the end of the horizontal spring that has a force constant k, oscillates with an amplitude A. When the spring is at its greatest extension and the block is instantaneously at rest, a second block of mass m2 is placed on top of it.
a) what is the smallest value for the coefficient of static friction mue(s) such that the second object does not slip on the first.
b) explain how the total mechanical energy E, the amplitude A, the angular frequency omega, and the period T of the system are affected by the placing of m2 on m1, assuming that the coefficient of friction is great enough to prevent slippage
a) what is the smallest value for the coefficient of static friction mue(s) such that the second object does not slip on the first.
b) explain how the total mechanical energy E, the amplitude A, the angular frequency omega, and the period T of the system are affected by the placing of m2 on m1, assuming that the coefficient of friction is great enough to prevent slippage

Explanation / Answer

When the mass m1 is at the maximum extension, it suffers the maximum acceleration it can possibly do. Also, all the energy of the system at this point of time is stored as the potential energy of the spring. Plus, when the block starts moving towards the equilibrium position, it starts to gain speed and the speed is maximum at the point of equilibrium. The total energy of the system, when at equilirbrium, is in the form of kinetic energy.

We will make use of the above to solve the given problems:

Part a.) Now, the force acting on the block at the maximum elongation would be kA

As both the blocks are to move towards the string, their acceleration would be kA / (m1 + m2)

Now this acceleration to the block m2 would be provided by the friction acting on it.

So, m1g = m1kA / (m1 + m2)

or, = kA / g(m1 + m2) is the required expression for the minimum coefficient of friction.

Part b.) As no external force is acting on the system, the total mechanical energy of the sytem will remain same as before, that is, equal to E

Again, as the total energy is to remain same, the amplitude will remain same as well: A

The time period of the system would change and would now be given as: 2((m1 + m2)/k)

Also, the new angular frequency would be given as: = (k / (m1 + m2))

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