In a truck-loading station at a post office, a small 0.200-kg package is released from...

a.) The kinetic energy of the package at point determines how much work is done by friction.
${\displaystyle {\mu }_{k}mgL=\frac{1}{2}m{v}_B^2}$ 
Then solve for ${\displaystyle {\mu }_k}$ 
${\displaystyle {\mu }_k=\frac{v_B^2}{2gL}=\frac{{\left(4.8\right)}^2}{2\left(9.8\right)\left(3\right)}=0.392}$ 
b) 
${\displaystyle W_{\text{other}}=K_B-U_A}$ 
${\displaystyle W_{\text{other}}=\frac{1}{2}m{v}^2-mgh}$ 
${\displaystyle =\frac{1}{2}\left(0.2\right){\left(4.8\right)}^2-\left(0.2\right)\left(9.8\right)\left(1.6\right)}$ 
${\displaystyle =-0.832}$ J

a) Kinetic friction on horizontal surface is equal to friction force

$\mu Nd\cos (\theta )=\frac{1}{2}m(v_2^2-v_1^2)$

N=mg

Given

$m=0.200kg;d=3.00m;v_1=4.80m/s;v_2=0$

Therefore 

$\mu (0.200kg)\times 9.8\times 3\cos {180}^{\circ }=\frac{1}{2}(0.200)\times (0^2-{4.80}^2)$

Step 2

b) Total work= work done by gravity + work done by friction

Total work = kinetic energy

$W_T=\frac{1}{2}m(v_1^2-u^2)$

$W_g+W_f=\frac{1}{2}m(v_1^2-u^2)$

$W_f=\frac{1}{2}m(v_1^2-u^2)-mgR=\frac{1}{2}0.200((4.80{)}^2-0))-(0.200)(9.8)(1.60)$

$W_f=-0.832J$ (work done against moving body , hence -ive sign)

Step 1

Given:

The mass is 2 kg

The radius is 1.60 m

The package's size is 1.6 m

The speed is given as 4.80 m/s

The distance is 3 m

Step 2

a) At point B, kinetic energy is equal to the friction's work done. The formula is denoted as,

$\frac{1}{2}m{v}_B^2={\mu }_{k}mg\times Bc$

${\mu }_k=\frac{v_B^2}{2g\times Bc}=\frac{{4.8}^2}{2\times 9.81\times 3}$

${\mu }_k=0.39$

Step 3

b) The sum of kinetic energy at point B and work done by the friction is equal to the Potential energy at point A.

The formula is denoted as,

$mgR=W_{F_{AB}}+\frac{1}{2}m{v}_B^2$

$W_{F_{AB}}=m(gR-\frac{1}{2}{v}_B^2)$

On substituting the values,

$W_{F_{AB}}=0.2(9.81\times 1.60-0.5\times {4.8}^2)$

$W_{F_{AB}}=0.835J$

Step 4

Solution:

a) At the horizontal surface, the coefficient of friction is 0.39

b) From A to B, the friction's work done is 0.835 J