Question #116285

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

Expert's answer

"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"

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