need help with work and energy

[ballahboy]

Two blocks, A(50kg) and B(100kg) are connected to a string. The pulley is frictionless and of negligible mass. The coefficient of kinetic friction between block A and the incline is Mk =0.25. Determine the change in the kinetic energy of block A as it moves from C to D, a distance of 20m up the incline if the system starts at rest.

What i did was i found Wg (work of gravitaional force), Wf (work of force) and Wff (work of frictional force) so i can use change of KE=Wg+Wf+Wff. For Wg i got -5897.79J, Wf = 19600J and Wff = -1956J. When i add these together, i got 11746.21J but the answer is suppose to be 3900J. What did i do wrong or do you guys prefer a different method.

Another question: A spring of length 080m rests along a frictionless 30degree incline. A 2.0kg mass, at rest against the end of the spring, compresses the spring by 0.10m. A) Determine the spring constant k . B) The mass is pushed down, compressing the spring an addtional 0.60m, and then released. If the incline is 2.0m long, determine how far beyond the rightmost edge of the incline the mass lands.

If someone could show me the steps to these problems, it would be greatly appreciated. Thanks!!

[ballahboy]

nvm i got the answer for the first one using a different formula..
can someone help me on the second one?

[Leong]

2.(a) static equilibrium. the sum of the forces which are parallel to the incline equals zero.

[ballahboy]

k used KEf+PEf+PEsf=KEi+PEi+PEsi to find the constant k and i got 3920N/M. Does that look right to ne one?

wuts static equilibrium? i dont think i learned that yet

[Leong]

i got 98.1 N/m. i don't know how to use the law of conservation of energy to get the answer, your answer.

u learned newton's 2nd law ?
a = 0 when the particle is not moving or moves with constant velocity. then the sum of the forces acting on the particle equals zero.

x axis is chosen to be parallel to the incline and positive in the right direction.
consider all the forces acting on the block and use newton's 2nd law :
-mgcos60+(-kx)=0
x=-0.10 m because the displacement of the spring is to the left hand side(when it is compressed) which is negative relative to its initial postion when it is relax.

[ballahboy]

thanks leong, i think i got part a now
does ne one kno how to do part b?