91Ó°ÊÓ

The acceleration or the deceleration of the elevator is, $\mathrm{a}=2.4 \mathrm{m}/{\mathrm{s}}^2$.

The net force on the body is the vector sum of all the forces acting on the body. The force is also calculated as the product of mass and the acceleration of the body. The free-body diagram helps us to understand the forces and their direction acting on the body.

Draw the free-body diagram of a lamp according to given conditions. Apply Newton’s law for each FBD. Then by solving equations calculate unknown mass and tension.

Formulae:

$∑\mathrm{F}=\mathrm{ma}$

Here, F is force acting on the body, is mass and is the acceleration of the body.

When the elevator is descending the net force on it is in the downward direction. The forces on the lamp are tension in the cord in the upward direction and weight of the lamp in the downward direction. Since, the elevator is moving downward, the lamp is also moving downward. It implies that the net force is in the downward direction.

From free body diagram I, applying Newton’s law to the situation we get

$$\begin{gathered} \mathrm{mg}-\mathrm{T}=\mathrm{m}(-\mathrm{a}) \\ \mathrm{m}=\frac{\mathrm{T}}{\mathrm{g}+\mathrm{a}} \end{gathered}$$

We assume upward direction as positive. Substituting the given values in the above equation, we get

$$\begin{gathered} \mathrm{m}=\frac{89\mathrm{N}}{\left(9.8\mathrm{m}/{\mathrm{s}}^2+2.4\mathrm{m}/{\mathrm{s}}^2\right)} \\ =7.3\mathrm{kg} \end{gathered}$$

Therefore, the mass of the lamp is$7.3\mathrm{kg}$ .

When the elevator is ascending upwards, the net force on the lamp is in the upward direction. It implies that the tension in the cord is greater than the weight of the lamp.

From free body diagram II, and applying Newton’s second law to this situation, we get

$$\begin{gathered} \mathrm{T}-\mathrm{mg}=\mathrm{ma} \\ \mathrm{T}=\mathrm{m}\left(\mathrm{g}+\mathrm{a}\right) \end{gathered}$$

Substitute the given values in the equation to calculate the tension.

$$\begin{gathered} \text{T=7.3kg}\left(9.8\mathrm{m}/{\mathrm{s}}^2+2.4\mathrm{m}/{\mathrm{s}}^2\right) \\ =89\mathrm{N} \end{gathered}$$

Hence, the tension in the cord is $89\mathrm{N}$.