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Standard IX

Physics

Mathematical Formulation of second law

Two bodies ha...

Question

Two bodies have masses in the ratio $3 : 4$ . When a force is applied on the first body, it moves with an acceleration of $6 m / s^{2}$ . How much acceleration will the same force produce in the second body?

2.25 m / s^{2}

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4.5 m / s^{2}

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6 m / s^{2}

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5 m / s^{2}

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Solution

The correct option is B $4.5 m / s^{2}$
$F o r c e = m a s s \times a c c e l e r a t i o n .$
So, for a constant force, $m_{1} a_{1} = m_{2} a_{2}$
$⟹ a_{2} = \frac{m_{1}}{m_{2}} \times a_{1} = \frac{3}{4} \times 6 = 4.5 m / s^{2}$

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Two bodies have masses in the ratio 3:4. When a force is applied on the first body, it moves with an acceleration of 6 m/s2. How much acceleration will the same force produce in the second body?

The correct option is B 4.5 m/s2Force=mass×acceleration.So, for a constant force, m1a1=m2a2⟹a2=m1m2×a1=34×6=4.5m/s2

Two bodies have masses in the ratio $3 : 4$ . When a force is applied on the first body, it moves with an acceleration of $6 m / s^{2}$ . How much acceleration will the same force produce in the second body?

The correct option is B $4.5 m / s^{2}$
$F o r c e = m a s s \times a c c e l e r a t i o n .$
So, for a constant force, $m_{1} a_{1} = m_{2} a_{2}$
$⟹ a_{2} = \frac{m_{1}}{m_{2}} \times a_{1} = \frac{3}{4} \times 6 = 4.5 m / s^{2}$