Question

Two blocks are connected by a light string that passes over a frictionless pulley as in the figure below. The system is released from rest while m2 is on the floor and m1 is a distance h above the floor.

(a) Assuming m1 > m2, find an expression for the speed of m1 just as it reaches the floor. (Use any variable or symbol stated above along with the following as necessary: g.) vf = Incorrect: Your answer is incorrect. (b) Taking m1 = 6.6 kg, m2 = 3.7 kg, and h = 3.9 m, evaluate your answer to part (a). Correct: Your answer is correct. m/s (c) Find the speed of each block when m1 has fallen a distance of 2.0 m. 4.93 Incorrect: Your answer is incorrect.

Answer 1

First define your initial and final position. To make it easier, the initial position when the system is at rest. We will define y=0 at the position of the masses at rest. When the system is released, m1 drops by a distance d, m2 rises by the same distance d, and the spring is stretched by the same distance d as well, since they are all connected. Initially the system has no kinetic energy or potential energy. After the system is displaced a distance, d, after release the system has gained potential energy - both gravitational potential energy, PEg and Spring Potential energy, PEs. Thus,
E(initial) = E (final)
0 = PEg (for m1) + PEg (for m2) + PEs
0 = m1g(-d) + m2gd + 0.5kd^2 (FY1: d is negative for m1 because m1 being drops down)
Now, divide both sides by d
0 = -m1g + m2g + 0.5kd
Isolate for d
2g (m1 - m2) / k = d

When typing this in webassign, be mindful of the notation (i.e. the 1 in m1 and the 2 in m2 should be typed as subscripts).

Hope this helps! :)