Question

C. A 15 0-kg box has a rubber bottom. It is sliding on a horizontal sheet of ice with a 30.0 force applied to it. Coefficient of friction is 0.22 Determine: a. The force of friction on the box The acceleration of the box c. Later, the horizontal force is reduced to 20.0-N. Determine the acceleration of the box for this condition. d. Is this static of kinetic friction D. A block is pulled along a horizontal surface by a force of 12.0 N. The block has a mass of 6.00 kg and it accelerates at 1.00 m/s. Find the coefficient of kinetic friction for the interface between the block and the surface

Answer 1

(c) since, f_limit (limiting fraction) f_l = mu_0 N where N = mg = 15 times 9.8 (from figure) N = 147 N so, f_l = 0.22 times 147 (because mu = 0.22) f_l = 32.34 (a) since, f_l = 32.34 which is < F (= 30 N) so, force fraction is f_l = 30 N (b) since applied force is, F = 30 N wherever applied force < limiting fraction force then body is at rest so, a bar = 0 rightarrow acceleration (c) Now, f_l = mu_s N f_l = 0.22 times 147 f_l = 32.34 > F (= 20 N) so, body is at rest Hence, a bar = 0 \r\n (d) since body is at rest so, it is static fraction (d) given, m = 6 kg mu_k = ? F = 12 N a = 1 m/s^2 from figure N = mg = 6 times 9.8 = 58.8 N since, it is kinetic fraction.