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A 84-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 0.73 as. The elevator travels with this constant speed for 5.0 s, undergoes a uniform negative acceleration for 1.4 s, and then comes to rest.
What does the spring scale register During the first 0.80s of the elevator’s ascent?

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okpalawalter8

Answer:

[tex]SR=949.2N[/tex]

Explanation:

From the question we are told that:

Mass [tex]M=84kg[/tex]

Speed [tex]V=1.2m/s[/tex]

Acceleration Time [tex]t_a=0.73[/tex]

Constant speed Time[tex]t_s=5.0s[/tex]

Deceleration time [tex]t_d=1.4s[/tex]

Generally the equation for Acceleration is mathematically given by

[tex]a=\frac{v}{t}[/tex]

Therefore acceleration for the first 0.80 sec is

[tex]a=\frac{1.2}{0.80}[/tex]

[tex]a=1.5m/s^2[/tex]

Therefore

Spring Reading=Normal force -Reaction

[tex]SR=m(g+a)[/tex]

[tex]SR=84(9.8+1.5)[/tex]

[tex]SR=949.2N[/tex]

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