Question

In the configuration shown (10m incline for m1 and 8m vertical length for m2 with a spring under it, both masses are linked by a rope and a pulley at the top), the 52.0 N/m spring is unstretched, and the system is released from rest. The mass of the block on the incline is m1 = 24kg and the mass of the other block is m2 = 5kg. (Neglect the mass of the pulley.). If the coefficient of kinetic friction between m1 and the incline is 0.24, then how fast (in m/s) are the blocks moving when m1 has slid 0.90 m parallel to the incline (and the 5.00 kg block has gained 0.90 m in elevation)?

Answer 1

Initial heights of the blocks: m1 -> H; m2 -> h
Initial energy of the “two blocks” system is
E1 = m1•g•H+m2•g•h
When the blocks covered s=0.9 m the energy of ”two blocks+spring” system is
E2= m1•g(H-s•sinα) +m1•v²/2 {m2•g(h+s) +m2•v²/2 +ks²/2.
From the law of conservation of energy
E1-E2 =W(fr)
W(fr) =μ•m1•g•cosα•s.
m1•g•H+m2•g•h- m1•g(H-s•sinα) - m1•v²/2 {m2•g(h+s) - m2•v²/2 -ks²/2 = μ•m1•g•cosα•s.
v=sqrt[{ 2gs(m1•sinα-m2)-kx²-2•μ•m1•g•cosα•s}/(m1+m2)]
I got v= 2.26 m/s. Check my calculations.

Answer 2

Sorry, how did you get the angle and what is x?