A block of mass 1.6kg is attached to a spring of force constant 1000N/m, The spring is compressed a distance -

Question

of 2cm and then the block is released from rest.(A)Cal the speed of the block as it passes through the equlibrium if a constant frictionless force of 4N retards the motion? (B)Find the speed of the block if it pass through the equilibrium for a frictionless surface ? (C) Cal the change in kinetic energy of the block ?

Answer 1

The energy of a spring is .5*k*x^2

Since the block returns to equilibrium, the work done by friction is
F*x

the net energy will be kinetic:
.5*m*v^2=.5*k*x^2-F*x
v=sqrt(k*x^2/m-2*F*x/m)

x is 2cm (2/100 m)
k=1000
m=1.6
F=4
Once v is calculated, the kinetic change is .5*m*v^2
j