Question

A 2.90 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0340 m . The spring has force constant 850 N/m . The coefficient of kinetic friction between the floor and the block is 0.42 . 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 0.0190 m from its initial position? (At this point the spring is compressed 0.0150 m .)

Answer 1

here,

mass , m = 2.9 kg

spring constant , K = 850 N/m

s = 0.019 m

uk = 0.42

x = 0.034 m

let the final speed of block be v

work done by friction = initial potential energy stored in the spring - final potential energy stored - kinetic energy gained

uk * m * g * s = 0.5 * k * x^2 - 0.5 * k * (x - s)^2 - 0.5 * m * v^2

0.42 * 2.9 * 9.81 * 0.019 = 0.5 * 850 * 0.034^2 - 0.5 * 850 * (0.034 - 0.019)^2 - 0.5 * 2.9 * v^2

solving for v

v = 0.34 m/s

the final speed gained is 0.34 m/s