What is the coefficient of kinetic friction between surface and block?

Question

A block of mass 1.37 kg is kept at rest as it compresses a horizontal massless spring (k = 84.7 N/m) by 6.27 cm. As the block is released, it travels 0.502 m on a rough horizontal surface before stopping.

The acceleration of gravity is 9.8 m/s^2 :

Calculate the coefficient of kinetic friction between surface and block.

Answer 1

Ok, here you have to use this :

The energy at the beginning minus the energy at the end, is the work made by the friction, in other words :

At the beginning, the energy is :

1/2*K*x^2, where x is 6.27 cm, and K is = 84.7 N/m

then : 1/2*84.7*(6.27/100)^2 = 0.16 Joules

The final energy is cero, because the block stops, so the energy will be :

Final energy = work made bye the force of friction.

0.16 = 1.37*9.8*u*distance

the distance = 0.502 and "u" is the coefficient of kinetic friction.

0.16 = 1.37*9.8*u*0.502

u = 0.02

Hope that might help you.