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Help me answer: - Laws of motion - JEE Main

A Particle of mass 0.3 kg  is subjected to a force F =-kx with k =15 N/m  What will be its initial acceleration if it is released from a point  20 cm away from the origin ?

  • Option 1)

    5\: m/s^{2}

  • Option 2)

    10\: m/s^{2}

  • Option 3)

    3\: m/s^{2}

  • Option 4)

    15\: m/s^{2}

 
Answers (1)
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As we learnt in

Spring force -

F= -Kx

- wherein

K=\left | \frac{F}{x} \right |= \frac{N}{m}

 

 

 

F=\: -\: k\: x

or \: \: \: F= -15\times \left ( \frac{20}{100} \right )= -3 N

Initial acceleration is over come by retarding force.

or \: \: \: m\times \left ( acceleration\: a \right )= 3

or \: \: \: a= \frac{3}{m}= \frac{3}{0.3}= 10ms^{-2}

Correct option is 2.


Option 1)

5\: m/s^{2}

This is an incorrect option.

Option 2)

10\: m/s^{2}

This is the correct option.

Option 3)

3\: m/s^{2}

This is an incorrect option.

Option 4)

15\: m/s^{2}

This is an incorrect option.

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