Two objects are connected by a light string passing over a light, frictionless p
ID: 2056653 • Letter: T
Question
Two objects are connected by a light string passing over a light, frictionless pulley. The 5.00?kg object is released from rest at a point 4.00 m above the floor. (a) Determine the speed of each object when the two pass each other. (b) Determine the speed of each object at the moment the 5.00?kg object hits the floor. (c) How much higher does the 3.00?kg object travel after the 5.00?kg object hits the floor?Explanation / Answer
(a) Apply conservation of energy to write an equation for each block separately, add them together, and solve for final velocity, v. For m1, it initially has some gravitational PE of m1gh and zero KE. When it arrives at a point even with m2, it has fallen half its original height of 4.00m, or 2.00m. At that time half it initial PE is now KE, therefore: m1gh = 0.5m1v² + m1gh / 2 2m1gh = m1v² + m1gh-------------->(1) The same can be said of m2, but it will have zero initial KE and PE, so that when it arrives even with m1 it will have both KE (of 0.5m2v²) and PE (of m2gh / 2): 0 = 0.5m2v² + m2gh / 2 0 = m2v² + m2gh-------------------->(2) Adding them together and solving for v: 2m1gh = (m1v² + m1gh) + (m2v² + m2gh) 2m1gh = m1v² + m2v² + m1gh + m2gh v²(m1 + m2) = m1gh - m2gh v² = gh(m1 - m2) / (m1 + m2) v = v[ gh(m1 - m2) / (m1 + m2)] = v[9.80m/s²(4.00m)(4.95kg - 2.50kg) / (4.95kg + 2.50kg)] = 3.95m/s (b) In (a), we considered a point half way up (2.00m), in this part h = 4.00m, so the equation we obtained in (a) would differ only in that it will have a coefficient of 2 on gh, so: v = v[ 2gh(m1 - m2) / (m1 + m2)] = v[2(9.80m/s²)(4.00m)(4.95kg - 2.50kg) / (4.95kg + 2.50kg)] = 5.078m/s (c) Because m2 is also moving at 5.078m/s when m1 hits the floor, it will have KE and assume zero PE at this point (we’re calling this point the zero point, and it will rise some point h above that point, so: mgh = 0.5mv² h = v² / 2g = (5.078m/s)² / 2(9.80m/s²) = 1.32m