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ed spring of length 0.15 m stands vertically on the floor; its stiffness is 1040 N/m. You release a block of mass 0.4 kg from rest, with the bottom of the block 0.7 m above the floor and straight above the spring. How long is the spring when the block comes momentarily to rest on the compressed spring

Respuesta :

samuelonum1

Answer:

0.082m

Explanation:

Step one:

given data

mass of block  m=0.4kg

initial length of spring=0.15m

spring constant= 1040N/m

height of mass= 0.7m

assume g=9.81m/s²

Step two:

When the mass is released it will have potential energy

PE=mgh-------1

when the mass comes to rest on the spring, the potential energy of the mass will be equal to the energy stored in the spring

the energy stored in the spring is given as

E=1/2kx²- -------------2

where e is the compression

mgh = mgh' + ½kx²

where h' = 0.15m - x = height of block at maximum spring compression

0 = mg(0.15m - x - 0.7m) + ½kx²

0 = 0.4* 9.81 * (-x - 0.55) + ½ * 1040 * x²

0=-3.924x-2.1582+ 520x²

rearrange

520x²-3.924x-2.1582=0

the roots are

x=0.068

x=-0.061----not possible

and x = 0.068 m  

so

h' = 0.15m - 0.068m = 0.082m

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