Question

A warehouse worker is pushing a 90.0 kg crate with a horizontal force of 284 N at a speed of v=0.850 m/s across the warehouse floor. He encounters a rough horizontal section of the floor that is 0.75 m long and where the coefficient of kinetic friction between the crate and floor is 0.351. (a) Determine the magnitude and direction of the net force acting on the crate while it is pushed over the rough section of the floor magnitude direction (b) Determine the net work done on the crate while it is pushed over the rough section of the floor. (c) Find the speed of the crate when it reaches the end of the rough surface. m/s terarum si An Viewing Save Wevert to Last Response

Answer 1

Given =90.0 kg 2840N V=00050 m/s d =0.75m u=0.351 using Newtow's law, we have mg F-f=ma Na 0 m f mg a) ung) - =-25-58 N こ Net force on the crate is given by Fnet F-f=(284N Fuet = (2 84 - 0.351x 90.0 X9:31) N (Fuerl 25.58 N Direction of force is left ward or opposite if the directon of motion. Net word done the fet force on the crate is given by Fuerida Fred d Cos 180° Wnet = ( 25.58 x0.75 X (-1)] I = -19.194 What = -19.191 b) by - Whet =

using aquation @ we have. F - f i ma ma a = 90° m F- umg -mg =( 284 - 0.351x9.8 0.351 x9.81)m/s a = -0.29 m/s? using equation ,v=u²tzas, we have, V²+zad V = [1.850) - 2x0.29% 9x0.75] 22 f = ms 0.2875 > = 0.536 m/s is 0.536 m/s Hence final speed