c. The block then starts to accelerate for another 4 seconds in the opposite direction with a magnitude of 2.5m/s. Compute the tension force from the chain within this duration d. Determine the final velocity of the block after the total 23 seconds and its vertical displacement from the starting point.
c. The block then starts to accelerate for another 4 seconds in the opposite direction with a magnitude of 2.5m/s. Compute the tension force from the chain within this duration d. Determine the final velocity of the block after the total 23 seconds and its vertical displacement from the starting point.
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please check the solution and solve part d
Transcribed Image Text: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/s
for 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 duration
c. The block then starts to accelerate for another 4 seconds in the opposite direction
with a magnitude of 2.5m/s". Compute the tension force from the chain within
this duration
d. Determine the final velocity of the block after the total 23 seconds and its vertical
displacement from the starting point.
Transcribed Image Text:Massf block (m) = 3000 kg.
Weight f block (W) =mg = (300)( 9. 8) = 29,4 30 NI
Weight of block acts in downward disection.
(a)
From Newtan's daw,
Bleck 13.5m-
Block
T-W=ma
T= Wt ma
T= 29,430 + (3000)(3.5)-
T=(29,430+ 10, So0) N
Tensjon force foom chain, T= 39,930 N
b) Since block iplifted by balance force, 80
Tension will be equal to weight.
So, T= W= 29,430 N
From Fuet =ma
Block
2. 5m/s?
W-T= ma
T= W-ma
T=29430-10 500
M.
T= 18930 N
Tension fosce foom chain,
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