Question

A 4.37 kg block free to move on a horizontal, frictionless surface is attached to one end of a light horizontal spring. The other end of the spring is fixed. The spring is compressed 0.117 m from equilibrium and is then released. The speed of the block is 1.01 m/s when it passes the equilibrium position of the spring. The same experiment is now repeated with the frictionless surface replaced by a surface for which uk = 0.345. Determine the speed of the block at the equilibrium position of the spring

Answer 1

given

m = 4.37 kg
x = 0.117 m
v = 1.01 m/s

let k is the spring constant.

Apply conservation of energy

(1/2)*k*x^2 = (1/2)*m*v^2

k*x^2 = m*v^2

k = m*v^2/x^2

= 4.37*1.01^2/0.117^2

= 325.6 N/m

on frictional surface where mue_k = 0.345

let v' is the speed of the block at equilibrium position.

Workdone by friction = change in mechanical energy

fk*x*cos(180) = (1/2)*m*v'^2 - (1/2)*k*x^2

-fk*x = (1/2)*m*v'^2 - (1/2)*k*x^2

-mue_k*m*g*x = (1/2)*m*v'^2 - (1/2)*k*x^2

-0.345*4.37*9.8*0.117 = (1/2)*4.37*v'^2 - (1/2)*325.6*0.117^2

==> v' = 0.478 m/s <<<<<<<<<<------------Answer