A block of mass 5 kg is placed at rest on a table of rough surface. Now, if a force of 30 N is applied in the direction parallel to surface of the table, the block slides through a distance of 50 m in an interval of time 10 s. Coefficient of kinetic friction is (given, $$g = 10$$ m s$$^{-2}$$):

We have a block of mass $$m = 5$$ kg with an applied force $$F = 30$$ N that travels a distance $$s = 50$$ m in time $$t = 10$$ s, starting from rest (with $$g = 10$$ m/s$$^2$$).

Since the block starts from rest, using $$s = ut + \frac{1}{2}at^2$$ gives

$$50 = 0 + \frac{1}{2} \times a \times (10)^2 = 50a$$

so $$a = 1$$ m/s$$^2$$.

Now applying Newton's second law along the surface (the friction force opposes motion):

$$F - \mu_k mg = ma$$

$$30 - \mu_k \times 5 \times 10 = 5 \times 1$$

$$30 - 50\mu_k = 5$$

$$50\mu_k = 25$$

So $$\mu_k = 0.50$$. Hence, the correct answer is 0.50.