In an Atwood machine, the heavier block has a mass twice that of the lighter block. When the system is released from rest, the tension in the string is 16.0 N . Determine the decrease in gravitational potential energy during the first second of motion.

Step-by-step solution

Let the mass of the lighter block be m and the heavier block be 2m . The acceleration of the system is a = 2m - m 2m + m g = g 3 . The tension in the string is T = 2(2m)(m) 2m + m g = 4mg 3 . Given T = 16.0 N , we have 4mg 3 = 16.0 mg = 12.0 N . The distance moved by the blocks in the first second ( t = 1 s ) is s = 1 2 at^2 = 1 2 ( g 3 )(1)^2 = g 6 . The decrease in gravitational potential energy is U = (2m)gs - mgs = mgs . Substituting the values, U = (12.0) ( g 6 ) = 2.0g . Taking g = 9.8 m/s ^2 , U = 2.0 9.8 =