Answer:
(A) Frictional force will be 103.7276 N
(B) Coefficient of friction [tex]\mu _k=0.4751[/tex]
Explanation:
We have given
mass of the box m = 25 kg
Angle [tex]\Theta =27^{\circ}[/text]
Acceleration [tex]a=0.3m/sec^2[/tex]
Apply Newton's second law of motion ,
Vertically :
[tex]F_N[/tex]= mg cosθ
= [tex]25\times 9.8\times COS27^{\circ}=218.296N[/tex]
Horizontally :
mg sinθ - F = ma
A)
frictional force F= mg sinθ - ma
= [tex]25\times 9.8sin27^{\circ}-25\times 0.3=103.7276N[/tex]
B)
Coefficient of friction [tex]\mu _k=\frac{F}{F_N}=\frac{103.7276}{218.296}=0.4751[/tex]
The frictional force impeding the motion of the box is 103.73 N
The coefficient of friction is 0.48
The given parameters;
The vertical component of the force on the box is calculated by applying Newton's second law of motion;
[tex]F_n = mg\times cos(\theta)\\\\F_n = 25\times 9.8 \times cos(27)\\\\F_n = 218.3 \ N[/tex]
The frictional force is determined by resolving the horizontal component of the force on the box;
[tex]\Sigma F_{||} = 0\\\\mgSin(\theta) - F_k = ma\\\\F_k = mg\times sin(\theta) - ma\\\\F_k = 25\times 9.8 \times sin(27) - (25\times 0.3)\\\\F_k = 103.73 \ N[/tex]
The coefficient of friction is calculate as;
[tex]\mu_k = \frac{F_k}{F_n} \\\\\mu_k = \frac{103.73}{218.3} \\\\\mu_k = 0.48[/tex]
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