The Work Energy Theorem for a Single Point Particle (a Single Object that can be
ID: 1459423 • Letter: T
Question
The Work Energy Theorem for a Single Point Particle (a Single Object that can be treated as a Point Particle). As shown below, a 2.000 kg mass is sliding down an inclined plane that makes an angle of 25.00 degrees as measured from the horizontal. The mass has an initial velocity of 3.000 m/s directed down the inclined plane, as shown. Initially there is no friction between the inclined plane and the mass. Using Newton's Laws, the forces acting on the mass were determined to be as shown below (you do not have to snow how this was done, or verify my results - I assume you could have done this by yourself by now and just did the work to save you some Exam time):Explanation / Answer
m =2 kg, N =17.8 N, FG =19.6 N, vi =3 m/s , theta =25degrees, d =0.4m
(a) WG =FG.d = FG*dcos(90-25) = [19.6*0.4*cos(65)] =3.3 J
5)] = -3.3 J
(b) WN =Ndcos(90) = 0
(c) W =WG +WN = 3.3 J
(d) Work energy thorem: Work = chnage in kinetic energy
W = Kf- kI =(1/2)mvf^2 - (1/2)mvi^2
vf^2 = 2W/m + vi^2 = [2*3.3/2] + (3*3) = 12.3
vf =3.51 m/s
(e) vf = 3.25m/s
Work energy thorem: Work = chnage in kinetic energy
W = Kf- kI
WG+WN+Wf = (1/2)mvf^2 - (1/2)mvi^2
3.3+0+Wf = [(1/2)(2*3.25*3.25)] - [(1/2)(2*3*3)]
3.3+Wf = 1.5625
Wf = 1.5625 - 3.3 = -1.7375 J
(f) Wf = Fk.d = (mu)N*d*Cos(0)
1.7375 = (mu) (17.8*0.4)
(mu) = 0.244
Coefficicent of kinetic friction between the mass and inclined plane is 0.244
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