Answer:
(a)v = √gd (b) Td - Td = 0 (c) -g/4 m/s²
Explanation:
Solution
Given that:
By the work energy theorem,
ΔKE = W = work done
1/2 *3m * v² + 1/2 * m * v² = 3mgd - mgd
so,
2 v² = 2gd
v = √gd
(b)W tension system = Td - Td = 0
(c) a = 3mg - mg/ 4m = g/2
Thus,
Acm = 3 m (g/2) + m (g/2)/4m
= -g/4 m/s²
Note: The complete question to this example is shown below
Two blocks, with masses m and 3m, are attached to the ends of a string with negligible mass that passes over a pulley, as shown above. The pulley has negligible mass and friction and is attached to the ceiling by a bracket. The blocks are simultaneously released from rest.
(a) Derive an equation for the speed v of the block of mass 3m after it falls a distance d in terms of m, d, and physical constants, as appropriate.
(b) Determine the work done by the string on the two-block system as each block moves a distance d.
(c) The acceleration of the center of mass of the blocks-string-pulley system has magnitude aCOM . Briefly explain, in terms of any external forces acting on the system, why aCOM
