On a platform of mass M = 20 kg, placed on a frictionless horizontal road, is a spring with elastic constant k = 1000 N/m and rest length L_0 = 0.5 m, which has one end attached to the platform and the other in contact with a mass M = 5 kg. There is friction between m and the platform, and the coefficient of kinetic friction is $\mu = 0.5$. Initially, the length of the spring is L = 0.2 m and it is released. Find: the velocities V_0 of the platform and v_0 of m when the spring reaches length L_0.
The problem is that I consider ∆l negative (because there is a compression of the spring), so ∆l = 0.2-0.5 = 0.3, whereas my professor considered it positive (0.5-0.2 = 0.3)
So we arrive at two different numerical results...
If I take W = Final Energy - Initial Energy, I have - \mu mg ∆l = ½ m v_0² + ½ M V_0² - ½ k ∆l², right?
@JohnRennie The spring is attached to a support placed on the right, so it moves to the left. I placed v_0 directed to the left and V_0 directed to the right...
No, we start with the spring compressed i.e. the initial state is with the spring compressed. Then we let go and the spring expands to the right pushing the block to the right.
@JohnRennie OK, so I have to equate g/k to gradient, yes?
@JohnRennie In this graph, if we put final lengths instead of elongations, the line does not pass through the origin. If we put elongations instead of final lengths, the line passes through the point (0,0), so it passes through the origin...
The documentation says: 1) determination of k_el from the relationship F = - k_el x and checking linearity... For what values of the elongations x of the springs is the linearity of Hooke's law not verifiable? Remark: the measurement of elongations is a relative measurement...