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4. The 75-lb block is released from rest 5 ft above the plate. Determine the compression of each spring when the block momentarily comes to rest after striking the plate. Neglect the mass of the plate. The springs are initially unstretched. * 0.25 ft k = 1000 Ib/t k = 1000 lb/ft %3D k' = 1500 lb/ft 0.580 and 0.303 ft O 0.580 and 0.330 ft O 0.508 and 0.330 ft O 0.503 and 0.380 ft

4. The 75-lb block is released from rest 5 ft above the plate. Determine the compression of each spring when the block momentarily comes to rest after striking the plate. Neglect the mass of the plate. The springs are initially unstretched. * 0.25 ft k = 1000 Ib/t k = 1000 lb/ft %3D k' = 1500 lb/ft 0.580 and 0.303 ft O 0.580 and 0.330 ft O 0.508 and 0.330 ft O 0.503 and 0.380 ft

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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4. The 75-lb block is released from rest 5 ft above the plate. Determine the compression of each spring when the block momentarily comes to rest after striking the plate. Neglect the mass of the plate. The springs are initially unstretched. * 5ft 0.25 ft EB k = 1000 Ib/t = 1000 Ib/ft k' = 1500 lb/ft 0.580 and 0.303 ft 0.580 and 0.330 ft 0.508 and 0.330 ft O 0.503 and 0.380 ft
Transcribed Image Text:4. The 75-lb block is released from rest 5 ft above the plate. Determine the compression of each spring when the block momentarily comes to rest after striking the plate. Neglect the mass of the plate. The springs are initially unstretched. * 5ft 0.25 ft EB k = 1000 Ib/t = 1000 Ib/ft k' = 1500 lb/ft 0.580 and 0.303 ft 0.580 and 0.330 ft 0.508 and 0.330 ft O 0.503 and 0.380 ft
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