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Standard IX

Physics

First Law of Motion

The masses of...

Question

The masses of $10 k g$ and $20 k g$ respectively are connected by a massless spring. A force of $200 N$ acts on the $20 k g$ mass. At the instant shown, the $10 k g$ mass has an acceleration of $12 m s^{- 2}$ , what is the acceleration of $20 k g$ mass ?

12 m s^{- 2}

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4 m s^{- 2}

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10 m s^{- 2}

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20 m s^{- 2}

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Solution

The correct option is B $4 m s^{- 2}$
Writing eq of motion, for 20kg mass

$200 - k x = 20 a_{2}$ ———-1
And for 10kg mass,

$k x = 10 a_{1}$ ———————-2

then, $k x = 10 * 12 = 120 N$ substitute this value in eq 1

$200 - 120 = 20 a_{2}$

$a_{2} = \frac{80}{20} = 4 m / s^{2}$

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The masses of 10kg and 20kg respectively are connected by a massless spring. A force of 200N acts on the 20kg mass. At the instant shown, the 10kg mass has an acceleration of 12ms−2, what is the acceleration of 20kg mass ?

The correct option is B 4ms−2Writing eq of motion, for 20kg mass200−kx=20a2 ———-1And for 10kg mass,kx=10a1 ———————-2then, kx=10∗12=120N substitute this value in eq 1200−120=20a2a2=8020=4m/s2

The masses of $10 k g$ and $20 k g$ respectively are connected by a massless spring. A force of $200 N$ acts on the $20 k g$ mass. At the instant shown, the $10 k g$ mass has an acceleration of $12 m s^{- 2}$ , what is the acceleration of $20 k g$ mass ?

The correct option is B $4 m s^{- 2}$
Writing eq of motion, for 20kg mass

$200 - k x = 20 a_{2}$ ———-1
And for 10kg mass,

$k x = 10 a_{1}$ ———————-2

then, $k x = 10 * 12 = 120 N$ substitute this value in eq 1

$200 - 120 = 20 a_{2}$

$a_{2} = \frac{80}{20} = 4 m / s^{2}$