What is the speed of the block when it has moved a distance of 0.0200 m from its initial position? (At this point the spring is compressed 0.0190 m). See details.

A 2.55 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0390 m. The spring has a force constant of 815 N/m. The coefficient of kinetic friction between the floor and the block is 0.35. The block and spring are released from rest and the block slides along the floor.

1 Answer
Apr 11, 2017

Answer:

#v=0.32" "m/s#

Explanation:

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#"When a spring is compressed by a Force of F,The potential energy "##"is stored on it."#

#"if the spring is released, the Potential energy turns into kinetic energy."#

#"The object is repulsed by the call-back force which occurs on the spring"#

#F'=K*Delta x#

#K=815 N/m#
#Delta x=0.0190" "m#

#F'=815*0.019=15.48" "N#

#F_f=k*m*g#

#F_f=0.35*2.55*9.81=8.76N" Friction force"#

#"acceleration of object is :"#

#a=(F'-F_f)/m#

#m=2.55kg" mas of object"#

#a=(15.48-8.76)/(2.55)#

#a=2.64" "m/s^2#

#v^2=2*a*displacement#

#v^2=2*2.64*0.0200#

#v^2=0.1056#

#v=sqrt(0.1056)#

#v=0.32" "m/s#