Question

You push horizontally a 0.60 kg block on a surface with friction. The two coefficients of friction are μ= 0.50 and μ-0.45. Given the magnitudes of the push force below, fill in the rest of the table. Leave blank any cell that has no value. F (N) N (N) fs (N) fk (N) a (m/s2) 4 5

Please show work on side on how you got the values.

Answer 1

Given: horizontally = 0.60 kg Two coefficient of friction are: mu_s = 0.50 mu_k = 0.45 Net force = mass*acceleration F - F_r = ma, F_r = mu N = mu mg implies E mu mg = ma where F_r is the frictional force, acting is static frictional force if body is moving, frictional force acting is kinetic frictional force N is the normal Reaction and it is same for all above \r\n N = mg = 0.6 kg * 9.81 m/s62 = 5.886 N N = 5.886 N For F = 1 N and F = 2N f_s = mu_s mg = 0.5* 0.6 kg* 9.81 m/s^2 f_s = 2.943 N static frictional force = 2.943, which is grater than applied force and the body will not move so acceleration is zero \r\n But F = 3N on words static friction is over come and the object is moving Now we have kinetic friction against motion f_k = mu_k mg = 0.45