solution with this problem please

Question

The block is released from rest at the position shown, figure 1. The coefficient of kinetic friction over length ab is 0.22, and over length bc is 0.16. Using the principle of work and energy, find the velocity with which the block passes position c. distance ab=3m and distance bc=2m

the block is positioned at an angle of 24 degress.

Answer 1

I'm making assumptions without being able to see Figure 1: a) the block stays on a ramp tilted from horizontal by 24°, and b) the block is at rest at point "a". If the latter is not true, you have to add some Kinetic Energy due to decrease in elevation from starting point to point "a".

Work done by friction, force * displacement, is F_f Δs = (μF_N) Δs, where magnitude of the normal force F_N equals the component of the weight perpendicular to the ramp: F_N = mg cos 24.

Over distance ab, Potential Energy change mgΔh goes into a change in KE and the work of friction, so mg(3 sin 24) = ΔKE₁ + (0.22)(mg cos 24)(3).

Over distance bc, mg(2 sin 24) -> ΔKE₂ + (0.16)(mg cos 24)(2).

Plugging in numbers, 1.220mg = ΔKE₁ + 0.603mg and 0.813mg = ΔKE₂ + 0.292mg.

Then total increase in KE is ΔKE₁ + ΔKE₂ = mg(0.617 + 0.521) = 1.138mg = 11.16m for g = 9.8.

Finally, if v = 0 at "a", then at "c", 1/2 m v² = 11.16m, mass m cancels, and v = 4.7 m/s.