# Question #d54fa

##### 1 Answer
Dec 25, 2016

$W \approx 1120 N$

#### Explanation:

The weight of an object with mass, $m$, which is located at or near the surface of the earth, is given by $W = m g$. When a person weighs themselves, the reading on the scale is actually the normal force exerted by the scale to support the person's weight. This is true as long as the person and the scale are stationary relative to each other. When the person and scale experience acceleration, however, things get a bit more complicated.

A basic force diagram for an object at rest on a level surface.

Our sum of forces statement depicting all the forces acting on the person looks like this:

$\sum \vec{F} = \vec{n} - {\vec{F}}_{g} = m \vec{a}$

Where $\vec{n}$ is the normal force and ${\vec{F}}_{g}$ is the force of gravity, and ${F}_{g} = m g$.

When the lift is at rest or moving at a constant velocity, there is no acceleration ($\vec{a} = 0$), and so in the situation where both the scale and person are stationary relative to each other, we have:

$\sum \vec{F} = \vec{n} - {\vec{F}}_{g} = m \left(0\right)$

$\implies \vec{n} - {\vec{F}}_{g} = 0$

Adding ${\vec{F}}_{g}$ to both sides,

$\implies \vec{n} = {\vec{F}}_{g}$

So, when the lift is stationary, the normal force exerted by the scale is equal to the force of gravity acting on the person, and thus the weight of the person is given by $\vec{n} = m g$. However, when we introduce acceleration, we no longer have that the sum of forces is equal to $0$.

We're given the weight of the man as $100 N$, which, as stated above, is equal to $m g$. This is the man's usual weight, when he isn't experiencing acceleration. We can find the mass of the man by dividing his weight by $g$, the constant for free-fall acceleration on Earth, given by $g = 9.8 \frac{m}{s} ^ 2$.

$W = m g$

$\implies m = \frac{W}{g}$

$\implies m = \frac{100 N}{9.8 \frac{m}{s} ^ 2}$

$\implies m = 10.2 k g$

We are looking for $\vec{n}$, the weight of the man as he experiences an acceleration of $100 \frac{m}{s} ^ 2$. We have:

$\vec{n} - {\vec{F}}_{g} = m \vec{a}$

$\implies \vec{n} = m \vec{a} + {\vec{F}}_{g}$

$\implies \vec{n} = m \vec{a} + m g$

$\implies n = \left(10.2 k g\right) \left(100 \frac{m}{s} ^ 2\right) + \left(10.2 k g\right) \left(9.8 \frac{m}{s} ^ 2\right)$

$\implies 1119.96 N \approx 1120 N$

Or, equivalently, $1.12 \times {10}^{3} N$

The closest option you have provided is c) $1500 N$.