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ki77a [65]
3 years ago
12

What will a spring scale read for the weight of a 58.0-kg woman in an elevator that moves (a) upward with constant speed 5.0 m/s

, (b) downward with constant speed 5.0 m/s, (c) with an upward acceleration 0.23 g, (d) with a downward acceleration 0.23 g, and (e) in free fall?
Physics
2 answers:
vazorg [7]3 years ago
8 0
<h2>Answer:</h2>

(a) 580 N

(b) 580 N

(c) 713.4 N

(d) 446.6 N

(e) 0 N [i.e no reading]

<h2>Explanation:</h2>

Newton's law states that the net force(∑F) acting on a body is the product of the mass (m) of the body and the acceleration (a) of motion of the body. That is;

∑F = m x a         --------------(i)

The net force is calculated as the algebraic sum of the upward forces and the downward forces on the body. i.e

∑F = total upward forces - total downward forces.

<em>Substitute the value of ∑F into equation (i) as follows;</em>

total upward forces - total downward forces = m x a         -----------(ii)

<em>From the question;</em>

Upward force =  the tension (T) of the scale's spring which determines the reading of the scale.

Downward force  = the weight (W) of the woman in the elevator

<em />

<em>Therefore putting this in equation (ii) gives;</em>

T - W = m x a           --------------------(iii)

<em>But;</em>

W =  m x g      [product of the mass (m) of the body and gravity (g = 10m/s²)]

<em>Therefore, equation (iii) becomes;</em>

T - (m x g) = m x a              -------------------(iv)

Now, taking upward motion as positive and downward motion as negative;

(a) The elevator moves upward with a constant speed of 5.0m/s. This means that the acceleration of the elevator is zero (0) though in the positive direction. i.e

a = 0;

<em>Substitute a = 0 into equation (iv) to get;</em>

T - (m x g) = m (0)

T = m x g         ----------------(v)

<em>Where;</em>

m = 58.0kg

g = 10m/s²

<em>Substitute these values into equation (v) as follows;</em>

T = 58.0 x 10 = 580N

Therefore, the reading on the scale will be 580N

(b) The elevator moves downward with a constant speed of 5.0m/s. This means that the acceleration of the elevator is zero (0) though in the negative direction. i.e

a = 0;

<em>Substitute a = 0 into equation (iv) to get;</em>

T - (m x g) = m (0)

T = m x g         ----------------(vi)

<em>Where;</em>

m = 58.0kg

g = 10m/s²

<em>Substitute these values into equation (vi) as follows;</em>

T = 58.0 x 10 = 580N

Therefore, the reading on the scale will be 580N

(c) The elevator moves with an upward acceleration of 0.23g. This means that the acceleration is +0.23g  [where g is taken as 10m/s²]

a = 0.23 x 10;

a = 2.3m/s²

<em>Substitute a = 2.3 into equation (iv) to get;</em>

T - (m x g) = m (2.3)

T - mg = 2.3m      ----------------(vii)

<em>Where;</em>

m = 58.0kg

g = 10m/s²

<em>Substitute these values into equation (vii) as follows;</em>

T - (58.0 x 10) = 58.0 x 2.3

T - 580 = 133.4

T = 580 + 133.4

T = 713.4N

Therefore, the reading on the scale will be 713.4N

(d) The elevator moves with a downward acceleration of 0.23g. This means that the acceleration is -0.23g  [where g is taken as 10m/s²]

a = -0.23 x 10;

a = -2.3m/s²

<em>Substitute a = -2.3 into equation (iv) to get;</em>

T - (m x g) = m (-2.3)

T - mg = -2.3m      ----------------(viii)

<em>Where;</em>

m = 58.0kg

g = 10m/s²

<em>Substitute these values into equation (viii) as follows;</em>

T - (58.0 x 10) = 58.0 x -2.3

T - 580 = -133.4

T = 580 - 133.4

T = 446.6N

Therefore, the reading on the scale will be 446.6N

(e) The elevator moves in free fall. This means that the acceleration is the one due to gravity (g) and because it is downwards, it is given as -g  [where g is taken as 10m/s²]

a = - 10;

a = -10m/s²

<em>Substitute a = -10 into equation (iv) to get;</em>

T - (m x g) = m (-2.3)

T - mg = -10m      ----------------(ix)

<em>Where;</em>

m = 58.0kg

g = 10m/s²

<em>Substitute these values into equation (ix) as follows;</em>

T - (58.0 x 10) = 58.0 x -10

T - 580 = -580

T = 580 - 580

T = 0N

Therefore, the reading on the scale will be 0N

Rasek [7]3 years ago
4 0

Answer:

(a) 58 kg

(b) 58 kg

(c) 71.34 kg

(d) 44.66 kg

(e) 0 kg

Explanation:

(a) & (b) when the elevator is moving at a constant speed, whatever the direction is, it would exert a 0 force and acceleration (according to Newton's 1st law) on the woman. Consequently the woman only return the same gravitational acceleration g on the scale.

(c) When the elevator is moving up with acceleration of 0.23g, the net acceleration returned by the woman is g + 0.23g = 1.23g. The scale would read 58 * 1.23 = 71.34 kg

(d) Similarly, when the elevator is moving down with acceleration of 0.23g, the net acceleration returned by the woman is g - 0.23g = 0.77g. The scale would read 58 * 0.77 = 44.66kg

(e) When all are in free fall, no one is exerting force on anything. Therefore the scale would experience no force on it at all, so it would read 0 kg.

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