------------------------------------ When an object slides on a surface, it enco
ID: 3341549 • Letter: #
Question
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When an object slides on a surface, it encounters a resistance force called friction. This force has a magnitude of muN, where mu is the coefficient of kinetic friction and N is the magnitude of the normal force that the surface applies to the object. Suppose an object of mass 30 kg is released from the top of an inclined plane that is inclined 30 deg to the horizontal (see Figure 3.11 in Section 3.4). Assume the gravitational force is constant, air resistance is negligible, and the coefficient of kinetic friction mu = 0.2. Determine the equation of motion for the object as it slides downt he plane. If the top surface of the plane is 5 m long, what is the velocity of the object; when it reaches the bottom?Explanation / Answer
From free body diagram,
N = mgcos30 = 30*9.8*cos30 = 254.611 N
=> uN = 0.2* 254.611 = 50.922 N
ma = mgsin30 - uN = 30*9.8*sin 30 - 50.922 => a=3.203 m/s2
L = 5 m => Final velocity = sqrt (2*3.203*5) = 5.66 m/s
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