Given :
A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º.
To Find :
The coefficient of static friction between the box and the plane.
Solution :
Vertical component of force :
Horizontal component of force(Normal reaction) :
Since, box is on the verge of slipping :
Therefore, the coefficient of static friction between the box and the plane is 1.07.
Hence, this is the required solution.
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Answer:
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Explanation:
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Wt. = Fg = m*g = 60kg * 9.8N/kg=588 N.=
<span>Wt. of skier. </span>
<span>Fp=588*sin35 = 337 N.=Force parallel to </span>
<span>incline. </span>
<span>Fv = 588*cos35 = 482 N. = Force perpendicular to incline. </span>
<span>Fk = u*Fv = 0.08 * 482 = 38.5 N. = Force </span>
<span>of kinetic friction. </span>
<span>d =h/sinA = 2.5/sin35 = 4.36 m. </span>
<span>Ek + Ep = Ekmax - Fk*d </span>
<span>Ek = Ekmax-Ep-Fk*d </span>
<span>Ek=0.5*60*12^2-588*2.5-38.5*4.36=2682 J. </span>
<span>Ek = 0.5m*V^2 = 2682 J. </span>
<span>30*V^2 = 2682 </span>
<span>V^2 = 89.4 </span>
<span>V = 9.5 m/s = Final velocity.</span>
