Solve part d . Masst block (m)= 3000 kg.Weight -f block (w)=mg = (300)(9. ) = 29,4 30 AI%3DWeight of block acts in downward disection.(a)From Newton's daw,Block3.5m/3T-W=maT= Wt maT= 29,430 +(3002) (3.5)T=(29,430+ 10, 5oo) NTension force foom chain, T= 39,930 N%3Db) Since block ip lifted by balance force, 80Tensjon will be equal to weight.So, T= W= 29,4 30 NFrom Fuet =ma| Block 2. 5m/sW-T= maT= W-maT=29430-10500M.T= 18930 NTension fosce foom chain, 6. A crane lifts a 3000kg block with a chain vertically upward.a. The resting block has been lifted upward with a uniform acceleration of 3.5m/sfor 4 seconds. Find the tension force from the chain.b. The block is now lifted with balanced vertical forces for another 15 seconds.Calculate the tension force from the chain within this durationc. The block then starts to accelerate for another 4 seconds in the opposite directionwith a magnitude of 2.5m/s. Compute the tension force from the chain withinthis duration.d. Determine the final velocity of the block after the total 23 seconds and its verticaldisplacement from the starting point.

Given data: Initial velocity for block u = 0 acceleration for time t = 0 to 4s is = 3.5 m/s2 for…

Answered: d. Determine the final velocity of the…

d. Determine the final velocity of the block after the total 23 seconds and its vertical displacement from the starting point.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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