The 2.0 kg block in the figure slides down a frictionless curved ramp, starting from rest at a height of h = 3.80 m.?

Take g = 9.81 m/s²

Consider the journey of sliding down the frictionless curved ramp.
Loss in P.E. = Gain in K.E.
m g h = (1/2) m v²
v = √(2 g h)
v = √(2 × 9.81 × 3.80) m/s

Consider the journey of sliding on the rough horizontal surface.
Initial speed, u = Final speed of sliding down the curved ramp = √(2 × 9.81 × 3.80) m/s

v² = u² + 2as
0 = (2 × 9.81 × 3.80) + 2 a (7.60)
Acceleration, a = -4.91 m/s²

F = m a
m g μ(k) = m a
μ(k) = a / g

Coefficient of friction = 4.91/9.81 = 0.5

Use conservation of energy to find the speed at the bottom of the ramp
PE initial = KE final
M*g*h = 1/2*m*v^2
V = sqrt (2*g*h)

Then use kinematics to find the deceleration
Vf^2 = vi^2 - 2*a*d

With vf = 0, Vi from above, solve for a
a = vi^2/(2*d) = 2*g*h /(2*d) = g*h/d

Then when you have the acceleration, simply use
Fnet = m*a

Since the only horizontal force is friction, the net horizontal force equals friction.

Friction = m*a

True definition of friction.
Friction = u*N

On the horizontal section, the only vertical forces are N and mg.

Sum of the vertical forces (up is positive ) = 0 = N - m*g.
N = m*g

So...
M*a = u*m*g

Solve for u
U = a/g = g*h/(g*d) = h/d

Now plug in numbers
U = 3.8 / 7.6 = 0.5

Hope the explanation helped