A 2.15kg block on a horizontal floor is attached to a horizontal spring that is

Q: A 2.15kg block on a horizontal floor is attached to a horizontal spring that is

Ei-Wf = Ef 0.5*k*A^2 - u*m*g*x1 = 0.5*m*v^2 + 0.5*k*x2^2 (0.5*825*0.04*0.04)-(0.39*2.15*0.019)=(0.5*2.15*v^2)+(0.5*825*0.021*0.021) v = 0.656 m/s

ID: 1381373 • Letter: A

Question

A 2.15kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0400m . The spring has force constant 825N/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.

Part A

What is the speed of the block when it has moved a distance of 0.0190m from its initial position? (At this point the spring is compressed 0.0210m .)

Express your answer with the appropriate units.

Explanation / Answer

Ei-Wf = Ef

0.5*k*A^2 - u*m*g*x1 = 0.5*m*v^2 + 0.5*k*x2^2

(0.5*825*0.04*0.04)-(0.39*2.15*0.019)=(0.5*2.15*v^2)+(0.5*825*0.021*0.021)

v = 0.656 m/s <-------answer

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A 2.15kg block on a horizontal floor is attached to a horizontal spring that is

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