Compression of a spring with friction

[azizlwl]

I agree. I measure the potential at various points of the journey with respect to the ground/ relative to the ground.

Maybe we approach different way.
If force is applied for a distance then the body gain energy equal to Fx
Now if gravity do work and PULL the object down, energy gain by the object.
The object gain energy depend on the distance it moved NOT from where(assuming g constant)

Does a point A higher than B have different ΔPE if moved equal distance?
Is position determine the energy equation?

The motion is between Point A, B and C only.
Making ground as reference will create unnecessarily messy calculation.

Add: You should also check about conservative force.

 

[PhanthomJay]

Many thanks. I think the reason I might have got confused is because azizlwl might have used a different reference of potential.
Ok, so using the velocity at C to be 15.7m/s, i have: $$\frac{1}{2}mv_c^2 + mgh_c = \frac{1}{2}kx^2 + mg\mu(CD)$$. Putting in numbers gives x ≈ 7.7m?
Where is my error?

Your error is that you forgot to include the gravitational potential energy term at E. It has the same value as the gravitational potential energy at C. Correcting for this oversight , you get x = 1 m. Same answer as the other approaches.

 

[CAF123]

Your error is that you forgot to include the gravitational potential energy term at E. It has the same value as the gravitational potential energy at C. Correcting for this oversight , you get x = 1 m. Same answer as the other approaches.

Thanks for pointing this out: all this discussion for a simple error like that. And thanks azizlwl - it is good to know there are other ways to solve the problem.