First, we need to find the resultant force acting on the block with the following equation:

F = F1(hori) + F2(hori) – μN

Where:

F1(hori) = F1 cos 20°, is the horizontal component of F1

F2(hori) = F2 cos 35°, is the horizontal component of F2

μN = μmg, is the force of friction.

So the total force on the body is:

F = F1 cos 20° + F2 cos 35° - μ(mg)

F = 100(0.94) + 75(0.82) - (0.4)(35)(9.8)

F = 18.2 N

With the value of the net force and the mass, we can calculate the acceleration experienced by the block applying Newton's second law:

F = ma

a = F/m = 18.2/35

a = 0.52 m/s^2

Now we can apply the equation of the distance to the uniformly accelerated motion:

d = v0t + a(t^2)/2

Then, for v0 = 0 and t = 10 s, we have:

d = a(t^2)/2 = (0.52)(100)/2

d = 26 m

In 10 seconds the block slides a distance of 26 m.

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