Question

A block of mass m is at rest at the origin at t = 0. It is pushed with constant force F_0 from x = 0 to x = L across a horizontal surface whose coefficient of kinetic friction is mu_k = mu_0 (1-x/L). That is, the coefficient of friction decreases from mu_0 at x = 0 to zero at x = L. a. Use what you've learned in calculus to prove that a_x = v_x dv_x/dx b. Find an expression for the block's speed as it reaches position L.

Answer 1

we know that a_x = dv_x/dt a_x = dv_x/dt dx/dx a_x = dv_x/dx dx/dt a_x = dv_x/dx v_x a_x = v_x dv_x/dx--(1) b) As the particle moving along x-axis we can find the net force acting on the block \r\n F_0 - f_k = ma F_0 - mu_k mg = ma F_0 - mu_0 [1 - x/L]mg = ma F_0/m - mu_0 (1 - x/L)g = a From 1 V_x dv_x/dx = F_0/m - mu_0 (1 - x/L)g v_x dv_x = F_0/m dx - mu_0 g dx + mu_0 g x/L dx integral_0^v_x v_x dx = F_0/m integral_0^L dx - mu_0g integral_0^L dx + mu_0s/L integral_0^L x dx v_x^2 = 2 F_0/m - 2mu_0 gL + mu_0 gL v_x = squareroot (2F_0/m - mu_0g)L