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3) A 60. kg person is in an elevator. The elevator starts from rest and then accelerates upwards at 2.0 m/s^2 for 4.0 seconds. Calculate the work done by the normal force on the person. *

Answer is 1.1 x 10^4 J

User Zistoloen
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Answer:

WD = 11328 J → 1.1 × 10⁴ (to 2 sig figs)

Step-by-step explanation:

WD = work done (J)

F = force (N), i.e. the normal force on the person

W = weight (N)

s = displacement (m)

m = mass (kg) = 60

a = acceleration (m/s²) = 2

g = gravity (m/s²) = 9.8

t = time (s) = 4

u = initial velocity (m/s) = 0

The formulas or equations that are relevant ate:

WD = F × s

F = m × a

W = m × g → note: this is simply a variation of the formula F = ma

s = ut + ¹/₂.at²

We want to find WD, so we need to know the force and the displacement (or distance);

Firstly, we want to find the normal force on the person in the elevator;

The downwards force the person produces on the elevator is the weight of the person, which is:

W = 60 × g

W = 60 × 9.8

W = 588

We are told the elevator is accelerating upward, so we can construct an equation to find F with the formula F = ma:

F - W = 60 × 2

F - 588 = 120

F = 708

Note: illustrations may be helpful to understand this, as has been shown in the picture

We also need displacement, which we use the formula s = ut + ¹/₂at²:

s = 0(4) + ¹/₂(2)(4)²

s = 16 m

Now we have F and s, we can calculate WD:

WD = 708 × 16

WD = 11328 J → This can be rounded to 1.1 × 10⁴ to 2 significant figures

3) A 60. kg person is in an elevator. The elevator starts from rest and then accelerates-example-1
User Joris Weimar
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