Certainly! Let's walk through the solution step-by-step.
We are given:
- The mass of the lump of meat, [tex]\( m = 5 \)[/tex] kg.
- The acceleration due to gravity, [tex]\( g = 10 \)[/tex] m/s².
### Part (i): When the elevator is moving with a steady speed
When the elevator is moving with a steady speed, it means there is no acceleration (i.e., [tex]\( a = 0 \)[/tex]). The only force acting on the mass is due to gravity.
The force (which is the reading on the spring balance) can be found using the equation:
[tex]\[ F = m \cdot g \][/tex]
Substituting the given values:
[tex]\[ F = 5 \, \text{kg} \times 10 \, \text{m/s}^2 \][/tex]
[tex]\[ F = 50 \, \text{N} \][/tex]
Therefore, the reading on the spring balance, when the elevator is moving with a steady speed, is:
[tex]\[ 50 \, \text{N} \][/tex]
### Part (ii): When the elevator is moving upwards with an acceleration of 0.2 m/s²
In this scenario, the elevator has an upward acceleration. This means that the total effective acceleration acting on the lump includes both the gravitational acceleration [tex]\( g \)[/tex] and the elevator's acceleration [tex]\( a \)[/tex].
The effective acceleration is:
[tex]\[ g + a \][/tex]
The force (which is the reading on the spring balance) can then be calculated using:
[tex]\[ F = m \cdot (g + a) \][/tex]
Substituting the values:
[tex]\[ F = 5 \, \text{kg} \times (10 \, \text{m/s}^2 + 0.2 \, \text{m/s}^2) \][/tex]
[tex]\[ F = 5 \, \text{kg} \times 10.2 \, \text{m/s}^2 \][/tex]
[tex]\[ F = 51 \, \text{N} \][/tex]
Therefore, the reading on the spring balance, when the elevator is moving upwards with an acceleration of 0.2 m/s², is:
[tex]\[ 51 \, \text{N} \][/tex]
In summary, the readings on the spring balance in the different scenarios are:
- Steady speed: [tex]\( 50 \, \text{N} \)[/tex]
- Accelerating upwards with [tex]\( 0.2 \, \text{m/s}^2 \)[/tex]: [tex]\( 51 \, \text{N} \)[/tex]
