annette1204
contestada

Assume that a scale is in an elevator on Earth.
what force would the scale exert on a 53-kg person
standing on it during the following situations?
(use Fnet = ma and F = mg and g = 10m/s^2)
a. The elevator moves up at a constant speed.
b. It slows at 2.0 m/s^2 while moving upward.
C. It speeds up at 2.0 m/s^2 while moving downward.
d. It moves downward at a constant speed.
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Assume that a scale is in an elevator on Earth what force would the scale exert on a 53kg person standing on it during the following situations use Fnet ma and class=

Respuesta :

BrainlyWarrior

★ Remember :

The reading of scale of a man in an elevator :

  • When the elevator moves upward with acceleration a, R = m(g + a)
  • When the elevator moves downwards with acceleration a , R = m(g - a)
  • When the elevator falls freely, we take a = g so, R = m(g - g) = 0
  • When the lift is at rest or moves with uniform velocity, a = 0, so R = m(g - 0) = mg

A). The elevator moves up at a constant speed[tex].[/tex]

∴ Acceleration of elevator = 0

➠ R = mg

➠ 53 × 10

➠ R = 530N

B). It slows at 2m/s² while moving upward.

∴ Acceleration of elevator = -2m/s²

[Negative sign shows that speed decreases with time]

➠ R = m(g + a)

➠ R = 53(10 + (-2))

➠ R = 53 × 8

➠ R = 424N

C). It speeds up at 2m/s² while moving downward.

∴ Acceleration of elevator = 2m/s²

➠ R = m(g - a)

➠ R = 53(10 - 2)

➠ R = 53 × 8

➠ R = 424N

D). It moves downward at constant speed.

∴ Acceleration of elevator = 0

➠ R = mg

➠ R = 53 × 10

➠ R = 530N

Hope It Helps!

onyebuchinnaji

(a) When the elevator moves up with a constant speed, the force on the scale is 519.4 N.

(b) When the elevator slows down at 2 m/s² while moving up,  the force on the scale is 625.4 N

(c) When the elevator speeds up at 2 m/s² while moving down, the force on the scale is 413.4 N.

(d) When the elevator moves down at constant speed, the force on the scale is 519.4 N.

The given parameters;

  • mass of the person , m = 53 kg

The force that scale must exert on the person is calculated as follows;

F = mg + ma

F = m(a + g)

(a)

When the elevator moves up with a constant speed, the acceleration is zero.

F = m(0 + g)

F = mg

F = 53 x 9.8 = 519.4 N

(b)

When the elevator slows down at 2 m/s² while moving up;

F = m(a + g)

F = 53(2 + 9.8)

F = 625.4 N

(c)

When the elevator speeds up at 2 m/s² while moving down;

F = m(g - a)

F = 53(9.8 - 2)

F = 413.4 N

(d)

when the elevator moves down at constant speed;

F = m(g - a)

F = 53(9.8 - 0)

F = 519.4 N

Learn more here:https://brainly.com/question/18801986

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