Consider that at t=0 (initially) m2 is H meters above the ground. In this problem, H=60 cm., R=7 cm., Mpulley=1.8 kg, m1=6 kg, and m2=15 kg. (a)Using those energies from the above table that apply to this question and identifying the ground as the point of "zero height", use "conservation of energy" to find the speed of m2 just as it hits the ground.
) Blocks of mass m1 and m2 are connected by a massless string that pulls over the pulley in the figure. Initially, the system is at rest. The table is frictionless. The pulley is frictionless, but it does have mass Mp and radius R. The moment of inertia of the pulley is Ipulley=2/3MR2 (note that this is DIFFERENT than what we have used in the practice test examples). Consider that at t=0 (initially) m2 is H meters above the ground. In this problem, H=60 cm., R=7 cm., Mpulley=1.8 kg, m1=6 kg, and m2=15 kg. (a)Using those energies from the above table that apply to this question and identifying the ground as the point of "zero height", use "conservation of energy" to find the speed of m2 just as it hits the ground. There are no non-conservative forces outside of the system to consider, so Ef=Ei (the initial state is when the system is at rest and m2 is hanging H meters above the ground, and the final state is when the system is moving (both blocks are moving and the pulley is spinning) just as m2 is hitting the ground). Remember that you can substitute for ? in terms of the speed "v" by using the formula for ? in the above table. Also, remember that h1 is not changing in this problem (so, therefore, mgh1 is not changing either), so it need not be considered in your final and initial energies. Also remember that the blocks are attached by a string, so they move with the same speed. Find the speed (in m/s) as m2 reaches the ground.
1) Blocks of mass m1 and m2 are connected by a massless string that pulls over the pulley in the figure. Initially, the system is at rest. The table is frictionless. The pulley is frictionless, but it does have mass Mp and radius R. The moment of inertia of the pulley is Ipulley=2/3MR2 (note that this is DIFFERENT than what we have used in the practice test examples). Consider that at t=0 (initially) m2 is H meters above the ground. In this problem, H=60 cm., R=7 cm., Mpulley=1.8 kg, m1=6 kg, and m2=15 kg.
(a)Using those energies from the above table that apply to this question and identifying the ground as the point of "zero height", use "conservation of energy" to find the speed of m2 just as it hits the ground. There are no non-conservative forces outside of the system to consider, so Ef=Ei (the initial state is when the system is at rest and m2 is hanging H meters above the ground, and the final state is when the system is moving (both blocks are moving and the pulley is spinning) just as m2 is hitting the ground). Remember that you can substitute for ? in terms of the speed "v" by using the formula for ? in the above table. Also, remember that h1 is not changing in this problem (so, therefore, mgh1 is not changing either), so it need not be considered in your final and initial energies. Also remember that the blocks are attached by a string, so they move with the same speed. Find the speed (in m/s) as m2 reaches the ground.
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