Answer to Question #116285 in Mechanics | Relativity for Balogun Raheem

Question

a 120n force applied horizontally drags a 4kg load along a horizontal table at a uniform velocity of 3m/s. What is the coefficient of kinetic friction between the load and the table

Answer 1

$m=4kg\\F=120N\\g=9.8\frac {m}{c^2}\\v=3\frac{m}{c}\\\mu-?\\Solution:\\\overline F+\overline{F_{fr}}+\overline{m\times g}+{ N}=0;\\-m\times g_y+N_y=0;\\N=m\times g\\F_x-F_{frx}=0;\\F_{frx}=F_x;F_{fr}=\mu\times N;\\F=\mu\times m\times g;\mu=\frac{F}{m\times g};\mu=\frac{120}{4\times9.8}=3.061\\Answer:\mu=3.061 \\$

there may be an error here and F=12N, then

$m=4kg\\F=12N\\g=9.8\frac {m}{c^2}\\v=3\frac{m}{c}\\\mu-?\\Solution:\\\overline F+\overline{F_{fr}}+\overline{m\times g}+{ N}=0;\\-m\times g_y+N_y=0;\\N=m\times g\\F_x-F_{frx}=0;\\F_{frx}=F_x;F_{fr}=\mu\times N;\\F=\mu\times m\times g;\mu=\frac{F}{m\times g};\mu=\frac{12}{4\times9.8}=0.3061\\Answer:\mu\approx0/3$