A 1 kg spring balance is attached to the ceiling of an elevator. An object A is suspended from the spring balance. When the elevator accelerates upwards at $2 m/s^2$, the reading on the spring balance is 7.2 kg. What will be the reading on the spring balance when the elevator decelerates downwards at $2 m/s^2$?

1. Problem Description

A 1 kg spring balance is attached to the ceiling of an elevator. An object A is suspended from the spring balance. When the elevator accelerates upwards at $2 m/s^2$, the reading on the spring balance is 7.2 kg. What will be the reading on the spring balance when the elevator decelerates downwards at $2 m/s^2$?

2. Solution Steps

Let $m_s$ be the mass of the spring balance, which is 1 kg.

Let $a$ be the acceleration of the elevator.

Let $m_A$ be the actual mass of the object A.

Let $g$ be the acceleration due to gravity, $g \approx 10 m/s^2$.

When the elevator accelerates upwards at $a = 2 m/s^2$, the reading on the spring balance is $m_1 = 7.2 kg$.

The equation for the apparent weight $W_1$ is given by:

$W_1 = m_A (g + a)$

Since the spring balance reads 7.2 kg, the force exerted by the spring balance is $m_1g = 7.2g$. Therefore,

$m_Ag(1 + \frac{a}{g}) = 7.2g$

$m_A (g + a) = m_1g$

$m_A (10 + 2) = 7.2 \times 10$

$12m_A = 72$

$m_A = \frac{72}{12} = 6 kg$

Now, when the elevator decelerates downwards at $a = 2 m/s^2$, it means the elevator is accelerating upwards at $a = -2 m/s^2$ or accelerating downwards at $a = 2 m/s^2$. The reading on the spring balance will be $m_2$.

The apparent weight $W_2$ is given by:

$W_2 = m_A (g - a)$

$m_2g = m_A (g - a)$

$m_2g = 6(10 - 2)$

$m_2g = 6(8)$

$m_2 \times 10 = 48$

$m_2 = \frac{48}{10} = 4.8 kg$

3. Final Answer

4.8 kg