Physics, asked by guptapriyanka7552, 10 hours ago

If mass attached to spring system made 4 times then what would be maximum velocity.

Answers

Answered by shreyas246
0

Initial total energy of the system = KE +PE

initially KE is zero and PE is (k∗x

2

)/2

as we know, maximum velocity will be attained when the block is at the equilibrium position because at equilibrium net force on the block will be zero.

now, when the block passes through the equilibrium position

we have,

PE is zero as there is neither expansion nor contraction in the spring

KE=(m∗v

2

)/2

applying conservation of energy gives us

(m∗v

2

)/2=(k∗x

2

)/2

v=(

(k/m)

)∗xInitial total energy of the system = KE +PE

initially KE is zero and PE is (k∗x

2

)/2

as we know, maximum velocity will be attained when the block is at the equilibrium position because at equilibrium net force on the block will be zero.

now, when the block passes through the equilibrium position

we have,

PE is zero as there is neither expansion nor contraction in the spring

KE=(m∗v

2

)/2

applying conservation of energy gives us

(m∗v

2

)/2=(k∗x

2

)/2

v=(

(k/m)

)∗x

Initial total energy of the system = KE +PE</h3><h3>initially KE is zero and PE is (k∗x2)/2</h3><h3>as we know, maximum velocity will be attained when the block is at the equilibrium position because at equilibrium net force on the block will be zero.</h3><h3>now, when the block passes through the equilibrium position</h3><h3>we have,</h3><h3>PE is zero as there is neither expansion nor contraction in the spring</h3><h3>KE=(m∗v2)/2</h3><h3>applying conservation of energy gives us</h3><h3>(m∗v2)/2=(k∗x2)/2</h3><h3>v=((k/m))∗x</h3><h3></h3><h3>

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