Suppose that your mass is 59.1 kg, and you are standing on a scale fastened to the floor of an elevator. The scale measures force and is calibrated in newtons. What does the scale read when the elevator is rising and its speed is decreasing at a rate of [tex]6.87 \, \text{m/s}^2[/tex]? Use [tex]9.80 \, \text{m/s}^2[/tex] for acceleration due to gravity.

The force that a scale in a rising and decelerating elevator reads can be calculated with the equation F= m*(g-a). Given that the mass, acceleration due to gravity, and elevator deceleration are 59.1 kg, 9.80 m/s2, and 6.87 m/s2 respectively, the force would be 172.8 Newtons.

Explanation:

The subject of this question is Physics, and it seeks to investigate the reading on a scale in an elevator when it is rising and decelerating. The force that the scale reads can be represented by the equation F= m*(g-a), where 'F' is the force, 'm' is the mass, 'g' is acceleration due to gravity, and 'a' is the rate at which the elevator is decelerating. In your case:

mass (m) = 59.1 kg
gravitational acceleration (g) = 9.80 m/s2
deceleration of the elevator (a) = 6.87 m/s2

So, the force the scale reads is F = 59.1 kg * (9.80 m/s2 - 6.87 m/s2). This is equal to 172.8 Newtons.

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calebtes calebtes · Answer 2 · 2024-05-31T04:31:42-04:00

Answer:

173.2 Newtons

Explanation:

Since the scale is a reading of the normal force, the force of gravity and the normal force are in opposite directions, which means the forces can be subtracted from each other. In this situation, we can use Newtons 2nd Law: Net Force =(mass)( acceleration). The net force in this situation would be the Force of Gravity minus the Normal Force. So the equation you will begin with is Fg(which is mg) - Normal Force = ma. We want the normal Force, so our new equation derived for Fnormal will be mg - ma = Fnormal.