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

1. Problem Description

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

2. Solution Steps

Let $m_A$ be the mass of object A.

When the elevator accelerates upwards at $2 m/s^2$, the scale reads 7.2 kg.

The reading on the scale is the tension $T$ in the spring, and it is given as $7.2 kg \times g$, where $g$ is the acceleration due to gravity, which is approximately $9.8 m/s^2$.

The free body diagram for object A includes the tension $T$ upwards and its weight $m_A g$ downwards. According to Newton's second law, $T - m_A g = m_A a$, where $a$ is the upward acceleration of the elevator.

Therefore, $T = m_A (g + a)$.

The spring scale also has mass $1kg$ so we have

$7.2 g = 1g + m_A (g + a)$

$7.2 g = 1g + m_A (g+2)$

$6.2 g = m_A (g + 2)$.

Assuming $g = 10 m/s^2$, $6.2(10) = m_A(10+2)$, thus, $62 = 12 m_A$. So $m_A = \frac{62}{12} = \frac{31}{6} kg$.

Now, when the elevator decelerates downwards at $2 m/s^2$, it is equivalent to accelerating upwards at $-2 m/s^2$. The acceleration $a = -2 m/s^2$.

The tension in the string, $T' = m_A (g+a) = m_A(g - 2)$.

Substituting $m_A = \frac{31}{6}$, we have $T' = \frac{31}{6} (10-2) = \frac{31}{6} (8) = \frac{31}{3} (4) = \frac{124}{3} N$.

The reading on the scale also includes the $1kg$ mass of the spring scale. Therefore $reading = \frac{T'}{g} + 1 = \frac{124}{3g} + 1$.

If we use $g=10$, $reading = \frac{124}{30} + 1 = \frac{124+30}{30} = \frac{154}{30} = \frac{77}{15} \approx 5.13 kg$.

Now let's consider the correct equations:

When elevator accelerates upwards with $2 m/s^2$ then

$T - m_A g - 1 \times g = (m_A + 1) a$, where $T$ is the tension read by the scale.

Given $T = 7.2g$, therefore $7.2g - m_A g - g = (m_A + 1) \times 2$, i.e.,

$6.2g = 2 m_A + 2$, $6.2(9.8) = 2 m_A + 2$, $60.76 = 2 m_A + 2$,

$2 m_A = 58.76$, $m_A = 29.38 kg$.

When elevator decelerates downwards, acceleration is $-2 m/s^2$.

$T' - m_A g - 1 \times g = (m_A + 1) (-2)$,

$T' - 29.38 g - g = -2(29.38 + 1)$,

$T' - 30.38 g = -2(30.38)$,

$T' = 30.38 g - 60.76$, $T' = 30.38 \times 9.8 - 60.76 = 297.724 - 60.76 = 236.964 N$.

Scale reading $= \frac{T'}{g} = \frac{236.964}{9.8} = 24.18 kg$. Since the question asks what is the reading, the reading is $\frac{T'}{g} = 24.18$. But the spring scale already has 1kg of mass. So apparent mass of $A$ would be $\frac{T'}{g} = 24.18$. Then the force equation of $A$ is:

$T' = m_Ag + m_A a = 29.38 \times 9.8 + 29.38 \times (-2) = 29.38 (9.8-2) = 29.38 (7.8)$. $T = 229.164 kg m/s^2$. So the scale reading is $5.91 kg$. The initial tension is $7.2 g$. $T=m_A g+m_A (2) + 1*g +1(2)$. Thus, $T=7.2g= m_A (g+2) +g +2$. so, $6.2 g - 2=m_A (11.8)$ and $m_A = 4.98$. Now the next equation:

$T_2 = m_A (g-2) +g-2$ $T_2= 4.98 * 7.8+9.8-2= 46.842$ scale rading $4.789 +1 = 5.789 \approx 4.8 kg$.

3. Final Answer

4.8 kg