A block of mass 2.5 kg moving with speed v on a horizontal surface collides with a horizontal spring of negligible mass and spring constant 320Nm^(-1) . The block compresses the spring by 8.5 cm from its equilibrium position. If the coefficient of friction between the block and the floor is 0.4. what is the speed of the block just before collision?

Find the energy stored by the spring, using the formula:

E_p = \frac{Fe}{2}

But you don't have the force! So, use the formula F = ke This changes the formula to:

E_p = \frac{ke^2}{2}

Then, using the principle of conservation of energy, the kinetic energy of the block just before it hits the spring is equal to the energy stored by the spring and the work done by the block due to friction.

Work done due to friction is given by W = \mu Rd

Kinetic energy is given by E_k = \frac12 mv_f^2

This gives:

\frac12 mv_f^2 = \mu Rd + \frac{ke^2}{2}

where:
m = mass of block
vf = velocity of block before collision
mu = coefficient of friction
R = normal force of the block (equal to the weight of the block)
d = distance moved by the block before it stops
k = spring constant
e = compression (also equal to distance moved, d)

Be careful for the units you use! Post what you get.

The only problem is I don't know if you have to consider the fact that the frictional force gradually decreases when the spring is compressed or not... but the way I did it, I just assumed it to be constant until it stops.

Erm.. thanks.. but where can I get for the Rd

erm..
is it the value of d is the compression of the spring, that is 8.5 cm ?

\mu this symbol is called mu. This is the coefficient of friction.

R is the normal reaction on the block. Gravity exerts a force mg on the block, and the surface on which the block is exerts this equal force, called the normal reaction denoted by 'R'. So, R = 2.5 x g. Where g is the acceleration due to gravity.
d is equal to the compression of the spring, that is 0.085 m.

OK.. thanks for the help