A rope of mass m uniformly distributed along it's length l can slide without friction on a horizontal table. At instant t=0 the hanging part of the rope has a length b, and the rope is held at rest. When it is left, the rope begins to slide. At instant t the part of the rope that hangs has a length x and a velocity v. Take the horizontal table as a reference.
Calculate the mechanical energy of the system at t=0 in terms of m,g,l,b
I guess potential energy is zero when entire rope lies on table :)
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Since no Joules are lost in this problem, the total mechanical energy remains ZERO
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That is the end but if you really want to make a problem:
is x not b ?
rho = mass/unit length = m/L
force pulling = rho x g = m g x/L
so
a = d^2x/dt^2 = g x/L
but for energy
change in potential =- mg(x/2)(x/L)
ke = (11/2) m v^2
find v by integrating a from x = 0 to x = b
To calculate the mechanical energy of the system at t=0, we need to consider the potential energy and the kinetic energy.
1. Potential Energy:
At t=0, the entire rope is at rest and hanging vertically. The potential energy of the rope is given by its height.
The height of the hanging part of the rope is (l - b) since b is the length of the rope on the table.
The potential energy (U) of the rope at t=0 can be calculated using the following formula:
U = m * g * h
where m is the mass of the rope, g is the acceleration due to gravity, and h is the height of the hanging part of the rope (l - b).
2. Kinetic Energy:
At t=0, the rope is held at rest, so it has no kinetic energy initially.
Therefore, the mechanical energy (E) of the system at t=0 is given by the potential energy:
E = U = m * g * (l - b)
So, the mechanical energy of the system at t=0 in terms of m, g, l, and b is m * g * (l - b).