A 2.7-kg block is hanging stationary from the end of a vertical spring that is attached...

A 2.7-kg block is hanging stationary from the end of a vertical spring that is attached to the ceiling. The elastic potential energy of the spring/mass system is 2.6 J. What is the elastic potential energy of the system when the 2.7-kg block is replaced by a 5.5-kg block?

Expert Solution

Given mass of the block initially m1 = 2.7 kg , elastic potential energy of spring mass system

E1 = 2.6 J

we know that when spring is hanging freely F= -kx = - mg ==> x= mg/k here k, m are same

x depends on mass only

elastic potential energy = 1/2 k x^2 = 2.6 J say E1 , E1 prportional to square of the elangation

now new energy E1/E2 = m1^2/m2^2

E2= E1*m2^2/m1^2

= 2.6*5.5^2 / 2.7^2

= 10.788 J

genius_generous answered 2 hours ago

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