Question

You are standing on a bathroom scale in an elevator in a tall building. Your mass is 64 kg . The elevator starts from rest and travels upward with a speed that varies with time according to$\mathbf{v}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\left(3.0m/s^2\right)\mathbf{t}\mathbf{+}\left(0.20m/s^3\right){\mathbf{t}}^{\mathbf{2}}$ . When t = 4.0 s , what is the reading on the bathroom scale?

Short answer

The reading on the bathroom scale is 94 kg.

Step-by-step solution

Identification of the given data

The given data can be listed below as:

• The mass of my body is m = 64 kg .
• The time, when the reading is taken, is t = 4.0 s .

Significance of the Newton’s second law

Newton’s second law states that the force exerted on an object, with constant mass, is directly proportional to the acceleration of that object. The second law mainly describes the motion of an object due to the involvement of force.

Determination of the reading of the scale

The free body diagram of the elevator has been drawn below:

From the above diagram, a normal reaction force R is acting in the upwards direction and the weight w is acting in the downwards direction. The acceleration is also in the upwards direction.

The equation of the velocity of the elevator is given as:

$\mathrm{v}\left(\mathrm{t}\right)=\left(3.0\mathrm{m}/{\mathrm{s}}^2\right)\mathrm{t}+\left(0.20\mathrm{m}/{\mathrm{s}}^3\right){\mathrm{t}}^2$

Here, v (t) is the velocity of the elevator at time t .

Differentiating the above equation with respect to the time t, we get acceleration-

$$\begin{gathered} \mathrm{a}\left(\mathrm{t}\right)=\frac{\mathrm{dv}\left(\mathrm{t}\right)}{\mathrm{dt}} \\ =\frac{\mathrm{d}\left(\left(3.0\mathrm{m}/{\mathrm{s}}^2\right)\mathrm{t}+\left(0.20\mathrm{m}/{\mathrm{s}}^3\right){\mathrm{t}}^2\right)}{\mathrm{dt}} \\ =\left(3.0\mathrm{m}/{\mathrm{s}}^2\right)+\left(0.4\mathrm{m}/{\mathrm{s}}^2\right)\mathrm{t} \end{gathered}$$

Here, a (t) is the acceleration at time t.

From the free body diagram, the net force on the block is given as-

$$\begin{gathered} \mathrm{R}-\mathrm{W}=\mathrm{ma}\left(\mathrm{t}\right) \\ \mathrm{R}=\mathrm{W}+\mathrm{ma}\left(\mathrm{t}\right) \end{gathered}$$

Here, is the normal reaction on the body, W is the weight of the elevator and m is my mass and a (t) is the acceleration with respect to the time .

Substitute mg for W and role="math" localid="1663773210229" $\left(3.0\mathrm{m}/{\mathrm{s}}^2\right)+\left(0.4\mathrm{m}/{\mathrm{s}}^2\right)\mathrm{t}\mathrm{for}\mathrm{a}\left(\mathrm{t}\right)$ in the above equation.

$$\begin{gathered} \mathrm{R}=\mathrm{mg}+\mathrm{m}\left[\left(3.0\mathrm{m}/{\mathrm{s}}^2\right)+\left(0.4\mathrm{m}/{\mathrm{s}}^2\right)\mathrm{t}\right] \\ =\mathrm{mg}\left[\mathrm{g}+\left(3.0\mathrm{m}/{\mathrm{s}}^2\right)+\left(0.4\mathrm{m}/{\mathrm{s}}^2\right)\mathrm{t}\right] \end{gathered}$$

Substitute the given values in the above equation.

$$\begin{gathered} \mathrm{R}=\left(64\mathrm{kg}\right)\left[9.8\mathrm{m}/{\mathrm{s}}^2+\left(3.0\mathrm{m}/{\mathrm{s}}^2\right)+\left(0.4\mathrm{m}/{\mathrm{s}}^2\right)\left(4.0\mathrm{s}\right)\right] \\ =\left(64\mathrm{kg}\right)\left[9.8\mathrm{m}/{\mathrm{s}}^2+\left(3.0\mathrm{m}/{\mathrm{s}}^2\right)+\left(1.6\mathrm{m}/{\mathrm{s}}^2\right)\right] \\ =921.6\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

On further solving, we get-

$$\begin{gathered} \mathrm{R}=921.6\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\times \frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2} \\ =921.6\mathrm{N} \end{gathered}$$

The reading of the bathroom scale M will be-

$$\begin{gathered} \mathrm{M}=\frac{\mathrm{R}}{9.8\mathrm{m}/{\mathrm{s}}^2} \\ =\frac{921.6\mathrm{N}}{9.8\mathrm{m}/{\mathrm{s}}^2} \\ =94\mathrm{kg} \end{gathered}$$

The bathroom scale will give the reading: 94 kg .