A mass m is hanging on a string that passes over a pulley and is then attached to another mass 3m that is resting on a horizontal table (along with the pulley). Neglect friction. Mass m is held montionless and is then released. When it has fallen a distance h, it will have a speed v which can be calculated from the formula

(a) v=sqrt((gh)/4)
(b) v=sqrt((gh)/2)
(c) v=sqrt(gh)
(d) v=sqrt(2gh)
(e) None of these are correct.

To solve this problem, we can use the principle of conservation of mechanical energy. Initially, the system has gravitational potential energy due to the height h (mgh), and when it reaches the bottom, it will have kinetic energy equal to the sum of the two masses (0.5 * (3m + m) * v^2).

According to the conservation of mechanical energy, the initial potential energy will be equal to the final kinetic energy:

mgh = 0.5 * (4m) * v^2

Simplifying the equation, we have:

gh = 2v^2

To find v, we can take the square root of both sides of the equation:

sqrt(gh) = sqrt(2v^2)

sqrt(gh) = v * sqrt(2)

Therefore, the correct answer is (e) None of these are correct. The correct formula for the speed v is v = sqrt(2gh).