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A block of m mass comes on a spring (constant k) from a height h. Below the spring is a block of mass M. What should be the value of m so that M lifts up the ground?

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3 years ago

Arun
25768 Points
```							Let the spring is elongated by a length x at equilibrium condition when a block of mass m is hanged from its end.Here force exerted by the weight of the block is F =mgAgain force constant of the spring being k,then by Hook's law F = kxSo kx = mgNow in this equilibrium state before application of sharp blow the PE of the spting-block system is = 1/2 k x^2The inital downward velocity of the suspended block is v.This means it has gained a KE = 1/2 m *v^2So initial toral mechanical energy just after the application of blow isE1 = 1/2 k *x^2 +1/2 m*v^2Now let the maximum elongation from equilibrium position be y after the application of sharp blow. Then in this conditon total elongation becomes x+y and the PE of the spring becomes = 1/2 k*(x+y)^2But the mass m has decreased its height by y.So there occurs a decrease in PE by the amount = mgy.When the block comes to an instantaneous rest its KE will be zero.By cinservation of energyWe get,y = v (m/k)^(1/2)
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3 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions