Calculating Stopping Distance with Spring Forces and Friction

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To calculate the stopping distance of a block sliding on a frictionless surface that encounters a spring and friction, one must consider the forces acting on the block. The initial kinetic energy of the block is converted into spring potential energy and work done against friction. The frictional force can be calculated using the coefficient of kinetic friction and the normal force, which is equal to the weight of the block. The stopping distance is determined by the point at which the spring force equals the frictional force, leading to the block's deceleration. Ultimately, the stopping distance is equal to the maximum compression of the spring, which can be derived from the energy balance involving kinetic energy, spring potential energy, and work done against friction.
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So I've been working on this for hours and trying to understand it but I just don't. It doesn't make sense to me. Here is my problem:

A block (M = 1.5 kg) is sliding along a frictionless surface with initial velocity Vo = 1.15 m/s. It comes in contact with the spring (k = 170 N/m) in the diagram, and when it does, it also experiences a friction force fk that opposed the motion. The coefficient of kinetic friction, (μk = 0.1).What is the stopping distance of the block?

The main thing is that I have no distance given or spring compression. Just Vo, k, M, and the coefficient of kinetic friction.
 
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The amount of spring compression is equal to the stopping distance.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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