If the unit of mass is doubled and the units of length an time are halved,how much is 20 N of force in new units? `(1" N= 1kg" m s^(-2))`

Options

A. 0.5 N
B. 5 new units of force
C. 5 N
D. 0.5 new units of force

Answer

To solve the problem, we need to analyze how the changes in the units of mass, length, and time affect the force measured in Newtons (N).

### Step-by-Step Solution:

1. **Understand the relationship of force**:
The formula for force (F) is given by:
\[ F = m \cdot a \]
where \( m \) is mass and \( a \) is acceleration. In SI units, 1 Newton (N) is defined as:
\[ 1 \, \text{N} = 1 \, \text{kg} \cdot \text{m/s}^2 \]

2. **Identify the changes in units**:
- The unit of mass is doubled: If the original mass is \( m \), the new mass will be \( 2m \).
- The unit of length is halved: If the original length is \( L \), the new length will be \( \frac{L}{2} \).
- The unit of time is halved: If the original time is \( T \), the new time will be \( \frac{T}{2} \).

3. **Express acceleration in terms of new units**:
Acceleration \( a \) is defined as:
\[ a = \frac{L}{T^2} \]
In the new units, the acceleration will be:
\[ a' = \frac{\frac{L}{2}}{\left(\frac{T}{2}\right)^2} = \frac{\frac{L}{2}}{\frac{T^2}{4}} = \frac{L}{2} \cdot \frac{4}{T^2} = \frac{2L}{T^2} \]

4. **Substitute the new mass and acceleration into the force equation**:
The new force \( F' \) can be calculated as:
\[ F' = \text{new mass} \cdot \text{new acceleration} = (2m) \cdot \left(\frac{2L}{T^2}\right) = \frac{4mL}{T^2} \]

5. **Relate the new force to the original force**:
The original force \( F \) was:
\[ F = m \cdot \frac{L}{T^2} \]
Thus, we can express the new force in terms of the original force:
\[ F' = 4 \cdot F \]

6. **Calculate the new force for the given value**:
Given that the original force is \( 20 \, \text{N} \):
\[ F' = 4 \cdot 20 \, \text{N} = 80 \, \text{N} \]

7. **Convert to new units**:
Since \( F' = 80 \, \text{N} \) in the new unit system, we need to find how it relates to the original units. We know that:
\[ F' = \frac{F}{4} \]
Therefore, in new units:
\[ 20 \, \text{N} = 4 \cdot F' \implies F' = \frac{20 \, \text{N}}{4} = 5 \, \text{N} \]

### Final Answer:
The force of \( 20 \, \text{N} \) in the new unit system is \( 5 \, \text{N} \).