Answer to Question #117500 in Mechanics | Relativity for Michael Mateus

Question

1. A force of 48.0 N is required to start a 6.0 Kg box moving across a horizontal
concrete floor.
a) What is the coefficient of static friction between box and floor?
b) If the 48.0 N force continues, the box accelerates at 0.70m/s2
, what is the
coefficient of kinetic friction?

Answer 1

Given data :

Force $F = 48N$

Mass of box $m = 6 Kg$

Normal Reaction $N=mg= 6 \times 9.81 N = 58.86 N$

Frictional Force $F_{Frictiona} = F = 48 N$

Static Friction $\mu_{Static} = \frac {F_{Friction}} {N}$

Static Friction:

$\mu_{Static} = \frac {48} {58.86} = 0.815$

Kinetic Friction :

$\mu_{Kinetic} = \frac {F-F_{Acceleration}} {N}$

$\mu_{Kinetic} = \frac {48 - 6\times 0.70 } {58.86} = 0.744$