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 ):

Step-by-step solution

Forces acting on block is shown below From the second equation of motion, s = 1 2 a t 2 ⇒ 50 = 1 2 a × 10 2 ⇒ a = 1   m   s - 2 Applying Newton's second law in vertical direction, R - m g = 0 ⇒ R = m g (where R is the magnitude of normal reaction on the block due to the surface) If μ is the coefficient of kinetic friction between the block and the surface, applying Newton's second law in horizontal direction, 30 - μ R = m a ⇒ 30 - μ m g = m a ⇒ 30