A 2.05 kg block on a horizontal floor is attached to a horizontal spring that is

Question

A 2.05 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0330 m. The spring has force constant 850 N/m. The coefficient of kinetic friction between the floor and the block is 0.38. The block and spring are released from rest and the block slides along the floor. What is the speed of the block when it has moved a distance of 0.0190 m from its initial position? (At this point the spring is compressed 0.0140 m) Express your answer with the appropriate units.

Explanation / Answer

initially block is at rest and spring is compressed.

initialKE = 0

here image is not that clear if value of spring constant is not 850 then change accordingly.

initial spring PE = kx^2 /2 = 850 ( 0.033^2) /2 = 0.463 J

finallly:

x = 0.014 m

spring PE = (850)(0.014^2)/2 = 0.0833 J

Work done by friction = - uk mg d = - (0.38 x 2.05 x 9.8 x 0.019)

Wf = - 0.145 J

Using work energy theorem,

work done by spring + work done by friction = change in KE

(0.463 - 0.0833)   - 0.145   = 2.05v^2 / 2 - 0

v = 0.48 m/s

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