A spring toy weighing 1 kg on a spring balance suddenly jumps upward. A boy standing near the toy notices that the scale of the balance reads 1.05 kg. In this process the maximum acceleration of the toy is - `(g = 10 m sec^(–2))`

Options

A. `0.05" m sec"^(–2)`
B. `0.5" m sec"^(–2)`
C. `1.05" m sec"^(–2)`
D. `1" m sec"^(–2)`

Answer

To solve the problem, we need to analyze the forces acting on the spring toy when it jumps upward and calculate the maximum acceleration.

### Step-by-Step Solution:

1. **Identify the Weight of the Toy**:
The weight (W) of the toy can be calculated using the formula:
\[ W = m \cdot g \]
where \( m = 1 \, \text{kg} \) (mass of the toy) and \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity).
\[ W = 1 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 10 \, \text{N} \]

2. **Determine the Reading on the Spring Balance**:
The reading on the spring balance when the toy jumps upward is given as 1.05 kg. To convert this reading into force (in Newtons), we use:
\[ F_s = \text{Reading} \cdot g = 1.05 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 10.5 \, \text{N} \]

3. **Set Up the Equation of Motion**:
When the toy jumps upward, the forces acting on it are the spring force (upward) and the weight (downward). The net force (F_net) acting on the toy can be expressed as:
\[ F_{\text{net}} = F_s - W \]
Substituting the values we calculated:
\[ F_{\text{net}} = 10.5 \, \text{N} - 10 \, \text{N} = 0.5 \, \text{N} \]

4. **Apply Newton's Second Law**:
According to Newton's second law, the net force is also equal to the mass of the toy multiplied by its acceleration (a):
\[ F_{\text{net}} = m \cdot a \]
Rearranging for acceleration gives:
\[ a = \frac{F_{\text{net}}}{m} \]
Substituting the values:
\[ a = \frac{0.5 \, \text{N}}{1 \, \text{kg}} = 0.5 \, \text{m/s}^2 \]

5. **Conclusion**:
The maximum acceleration of the toy is:
\[ a = 0.5 \, \text{m/s}^2 \]

### Final Answer:
The maximum acceleration of the toy is \( 0.5 \, \text{m/s}^2 \).

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