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A 72-kg man stands on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.2 m/s in 0.80 s. The elevator travels with this constant speed for 5.0 s, undergoes a uniform negative acceleration for 1.5 s, and then comes to rest. What does the spring scale register During the first 0.80s of the elevator’s ascent?

Sagot :

Answer:

Explanation:

During the first .8 s , the elevator is under acceleration . It starts from initial velocity u = 0 , final velocity v = 1.2 m /s , time = .8 s

v = u + at

1.2 = 0 +  .8 a

a = 1.2 / .8

= 1.5 m /s²

During the acceleration in upward direction , let reaction force of ground on man be R .

Net force on man = R - mg

Applying Newton's 2 nd law

R - mg = ma

R = m ( g + a )

= 72 ( 9.8 + 1.5 )

= 813.6 N .

This reaction force will be measured by spring scale , so reading of spring scale will be 813.6 N .

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