In the figure below the spring (k = 75 n/m) has an equilibrium length of 1.00 m. the spring is compressed to a length of 0.50 m and a mass of 2.0 kg is placed at its free end on a frictionless slope. the spring is then released. (a) if the mass is not attached to the spring, how far up the slope will the mass move before coming to rest? (b) if the mass is attached to the spring, how far up the slope will the mass move before coming to rest? (c) suppose now that the has a coefficient of kinetic friction uk. if the block, ttached to the spring, is observed to stop just as it reaches the spring's equilibrium position, what is the coefficient of friction uk.?

Now that we have a feel for the state of the circuit in its steady state, let us obtain the expression for the current in the circuit as a function of time. note that we can use the loop rule (going around counterclockwise): e−vr−vl=0. note as well that vr=ir and vl=ldidt. using these equations, we can get, after some rearranging of the variables and making the subsitution x=er−i, dxx=−rldt. integrating both sides of this equation yields x=x0e−rt/l. use this last expression to obtain an expression for i(t). remember that x=er−i and that i0=i(0)=0. express your answer in terms of e, r, and l. you may or may not need all these variables. use the notation exp(x) for ex.