Sliding penny

Frictional force can be expressed as

$Ff= μ · N$

$N = m · g$

where

$Ff = frictional$ $force (N, lb)$

$μ = frictional$ $coefficient$ $static$ $or$ $kinetic$ $(μs, μk)$

$N = normal$ $force$ $(N, lb)$

$m = mass$ $of$ $the$ $object$ $(kg)$

$g = acceleration$ $of$ $gravity$ $(9.8 m/s^{2})$

Input the number..

$N = 0.003 kg · 9.8m/s^{2}$

$N = 0.0294 N$

$0.0147 N = μ · 0.0294 N$

$μ =0.5$

The formula for the Coefficient of Friction is $\mu = \frac{F_f}{F_N}$ where:

• $F_f =$ Force of Friction, and

• $F_N$ is the Normal force.

Since the penny is sliding horizontally (perpendicular to the Force of Gravity), $F_N = F_g = ma$

Using mass in kg, and g as 9.8, we find:

$\mu = \frac{0.0147} {0.003 \times 9.8} = 0.5$

frictional force=coefficient of friction * normal reaction.

The coefficient of friction ( $u$ ), can be caluclated using the equation:

$u = \frac{F_f}{m \cdot g}$

Using the given values, we get:

$u = \frac{0.0147}{0.003 \cdot 9.8} = \fbox{0.5}$

f=uR u= coefficient of friction R=reaction force =mg m=0.003 kg g=9.8 m/s sq. 0.0147=u * 0.003 * 9.8 u=0.5

The formula for friction is $f = kN,$ where $k$ is the desired coefficient of friction and $N$ is the normal force acting on the object. In this case, the normal force is gravity, because the coin is sliding horizontally. Thus, the normal force has magnitude $0.003g$ newtons. Substituting, we get $0.0147 = 0.003g\cdot N.$ Solving for $N,$ we get $N \approx \boxed{0.5}.$

as coefficient of friction is [(frictional force exerted)/(weight of object)] So,[(0.0147)/(0.003*9.8)]=0.5

Since friction is kinetic , thus

u m g = F = 0.0147 N

m = 0.003 kg

Putting the values and on solving we get u = 0.5 which is the coefficient of kinetic friction

F=Nu(meu) u=mg 3*9.8=29.4 kgm/s2 F=Nu F/N=u 0.0147/29.4=0.5

Frictional force can be expressed as

Ff = μ · N        (1)

where

Ff = frictional force (N, lb)

μ = static (μs) or kinetic (μk) frictional coefficient

N = normal force (N, lb)

N=mg

so u=F/mg

u=0.0147/3 9.8 10^-3 = 0.5

The coefficient of friction equal 0.5