Scale in an elevator physics problem

In summary, a 62-kg girl weighs herself by standing on a scale in an elevator, and when the elevator is ascending at 11 m/s but its speed is decreasing by 5 m/s in each second, the scale reads 298.22 N.
  • #1
conniechiwa
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A 62-kg girl weighs herself by standing on a scale in an elevator. What does the scale read when the elevator is ascending at 11 m/s but its speed is decreasing by 5 m/s in each second?

I'm not really sure where to begin.
 
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  • #2
conniechiwa said:
A 62-kg girl weighs herself by standing on a scale in an elevator. What does the scale read when the elevator is ascending at 11 m/s but its speed is decreasing by 5 m/s in each second?

I'm not really sure where to begin.

Hint, you need the force exerted by the girl on the scale... that's the weight the scale reads...

You also know that the force the girl exerts on the scale = the force the scale exerts on the girl.
 
  • #3
Well I know that normally the scale would read 608.5 N if the elevator was not moving, but I'm not sure what to do here.
 
  • #4
conniechiwa said:
A 62-kg girl weighs herself by standing on a scale in an elevator. What does the scale read when the elevator is ascending at 11 m/s but its speed is decreasing by 5 m/s in each second?

I'm not really sure where to begin.

First, find what the deceleration of the elevator is.
 
  • #5
conniechiwa said:
Well I know that normally the scale would read 608.5 N if the elevator was not moving, but I'm not sure what to do here.

draw the freebody diagram of the girl... what are the forces acting on the girl... you know that [tex]\Sigma\vec{F} = ma[/tex]. follow l46kok's suggestion and find the acceleration of the girl.
 
  • #6
Would the deceleration be -3.45 m/s squared?
 
  • #7
conniechiwa said:
Would the deceleration be -3.45 m/s squared?

No, the deceleration is the same given in the question...
 
  • #8
only the velocity is given?
 
  • #9
conniechiwa said:
only the velocity is given?

They give 5m/s per second... so the acceleration is -5m/s^2
 
  • #10
iidesjo!

okay,

Fg=m*g (Fg is gravitational force)
Girl weighs (m) 62 [kg].
Gravitational constant (g) is 9.81 [m/s^2]
Fg= 62*9.81= 608.22 [N]
that is the static value.

With F=m*a you can calculate the difference in force compared to static (v=c)
a= -5 [m/s^2]
F= 62*-5= -310 [N]

so the scale will indicate 608.22-310= 298.22 [N]
to calculate back to mass divide through the graviotational constant.
298.22/9.81= 30.4 [Kg]
you're welcome!
:D
 

FAQ: Scale in an elevator physics problem

What is scale in an elevator physics problem?

Scale in an elevator physics problem refers to the numerical representation of the weight or force acting on an object in the elevator. It is usually measured in units of Newtons (N) or kilograms (kg).

How is scale related to acceleration in an elevator physics problem?

The scale in an elevator physics problem is directly related to the acceleration of the elevator. This means that as the elevator accelerates, the scale reading will also change to reflect the increase or decrease in force acting on the object.

How does the scale reading change when the elevator is stationary?

When the elevator is stationary, the scale reading will be equal to the weight of the object. This is because there is no acceleration acting on the object, so the force of gravity is the only force acting on it.

What happens to the scale reading when the elevator is moving at a constant velocity?

When the elevator is moving at a constant velocity, the scale reading will remain the same. This is because the object is not experiencing any change in acceleration, so the force acting on it remains constant.

How does the scale reading change when the elevator is moving up or down?

When the elevator is moving up, the scale reading will increase because the acceleration is acting in the same direction as the force of gravity. Conversely, when the elevator is moving down, the scale reading will decrease because the acceleration is acting in the opposite direction of the force of gravity.

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