A spring toy weighing $1. K g$ on a spring balance suddenly jumps upward. A boy standing near the toy notices that the scale of the maximum acceleration of the toy is $(g = 10 m {s e c}^{- 2})$

0.05 m {s e c}^{- 2}

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0.5 m {s e c}^{- 2}

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1.05 m {s e c}^{- 2}

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1 m {s e c}^{- 2}

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Solution

The correct option is B $0.5 m {s e c}^{- 2}$
According to the question as the toy jumps of the balance, the normal force acting on the balance is going to increase, which is 1.05 kg according to the question.
So, we know that initially when the toy was at rest the normal force was $m \times g = 10 N$ .
So, now the change in force due to the jump is
$= 1.05 g - 10$
$= 0.5 N .$
Wo know that, the accelaration is,
$a = \frac{F}{m} = \frac{0 \cdot 5}{1} = 0.5 m / s^{2}$

$∴$ The acceleration is $\to 0.5 m /$ $s^{2}$ (Ans).

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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 maximum acceleration of the toy is (g=10 m sec−2)

A spring toy weighing $1. K g$ on a spring balance suddenly jumps upward. A boy standing near the toy notices that the scale of the maximum acceleration of the toy is $(g = 10 m {s e c}^{- 2})$