A 2.60kg block on a horizontal floor is attached to a horizontal spring that is

Question

A 2.60kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0310m . The spring has force constant 810N/m . The coefficient of kinetic friction between the floor and the block is 0.39 . The block and spring are released from rest and the block slides along the floor.

A-What is the speed of the block when it has moved a distance of 0.0120m from its initial position? (At this point the spring is compressed 0.0190m .)

Express your answer with the appropriate units.

I only got one attempt remaining.

I tried.. (.4323 , .4276 , -.39 , -.62).. Thank you.

Explanation / Answer

energy is always conserved

intial spring energy = final spring energy + kinetic energy of block + work done by friction

0.5*810*0.0312 = 0.5*810*0.0192 + 0.5*2.6*V2 + 2.6*9.8*0.39*0.012

V= 0.308 m/s

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