Consider two blocks hung by a massless string from a frictionless pulley as shown in the figure.The masses of the blocks are m₁=5 kg and m₂=7 kg.The bottoms of the blocks are 20 meters above the top of the table.What is the speed of m₂ just before it strikes the table 20 meters below its starting point (within 5% accuracy)? A)22 m/s B)8.1 m/s C)20.m/s D)11.m/s E)1.7 m/s

Question

Quizplus · Accepted Answer

Since the pulley is frictionless, the tension in the string will be the same on both sides. Therefore, we can consider the two blocks as a single system. The total mass of the system is 5+7=12 kg. Using conservation of energy, we can relate the initial potential energy of the system to the final kinetic energy of the system before the second block hits the table:
mgh = (1/2)mv^2
where m is the total mass of the system, g is the gravitational acceleration, h is the initial height (20 m), v is the final velocity of the system before the second block hits the table.
Solving for v, we get:
v = sqrt(2gh)
Substituting the values, we get:
v = sqrt(2*9.8*20) = 19.8 m/s
Rounding off to nearest integer gives us the answer as 20 m/s, which is closest to option B (8.1 m/s) with an accuracy of 5%.

Consider Two Blocks Hung by a Massless String from a Frictionless