Two blocks of masses m = 3kg and M = 7kg are connected by a light rope that is swung over a smooth pulley which is hanging from the roof. Both blocks are initially at rest and at the same height of H = 2 meters from the floor. After the blocks are released, with what speed will block M strike the floor?
Question
Two blocks of masses m = 3kg and M = 7kg are connected by a light rope that is swung over a smooth pulley which is hanging from the roof. Both blocks are initially at rest and at the same height of H = 2 meters from the floor. After the blocks are released, with what speed will block M strike the floor?
Solution
To solve this problem, we will use the principle of conservation of energy and the concept of tension in a string.
Step 1: Identify the forces acting on the system The forces acting on the system are the gravitational force (weight) of the two blocks and the tension in the string.
Step 2: Apply the principle of conservation of energy The total mechanical energy of the system is conserved because there are no non-conservative forces doing work on the system. This means that the initial potential energy of the system (when the blocks are at rest at height H) is equal to the final kinetic energy of the system (when block M strikes the floor).
The initial potential energy of the system is the sum of the potential energies of the two blocks, which is mgH + MgH.
The final kinetic energy of the system is the sum of the kinetic energies of the two blocks, which is 1/2mv^2 + 1/2Mv^2.
Setting these two equal to each other, we get:
mgH + MgH = 1/2mv^2 + 1/2Mv^2
Step 3: Solve for v Rearranging the equation above, we get:
v^2 = 2gH*(m + M) / (m + M)
Solving for v, we get:
v = sqrt(2gH)
Substituting the given values, we get:
v = sqrt(29.82) = sqrt(39.2) = 6.26 m/s
Therefore, block M will strike the floor with a speed of 6.26 m/s.
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