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Grade: 11

                        

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?

3 years ago

Answers : (1)

Arun
24742 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 =mg

Again force constant of the spring being k,then by Hook's law F = kx

So 

kx = mg

Now in this equilibrium state before application of sharp blow the PE of the spting-block system is 

= 1/2 k x^2

The inital downward velocity of the suspended block is v.This means it has gained a KE 

= 1/2 m *v^2

So initial toral mechanical energy just after the application of blow is

E1 = 1/2 k *x^2 +1/2 m*v^2

Now 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)^2

But 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 energy

We get,

y = v (m/k)^(1/2)

3 years ago
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