2. A particle with mass m starts from point A and goes along track ABC according

Question

2. A particle with mass m starts from point A and goes along track ABC according to the figure. The plane of the track is vertical, segment AB is rough while BC is smooth. The radius R of segment BC and other data are given below Data R=2m, = 30° 7m As 4m, g-9,81 a, Calculate the work done by the gravitational and reaction forces on segment AB b, Write up the work-energy theorem on segment AB and calculate the magnitude of the velocity of the particle in point B c, Write up the impulse-momentum theorem on segment AB and calculate the elapsed time between positions A and B of the particle. d, Write up the work-energy theorem on segment BC and calculate the magnitude of the velocity of the particle in point C. f, Calculate the magnitude of the reaction force that acts on the particle in point C.

Explanation / Answer

a) Work done by reaction forces will be zero since displacement and reaction forces are perpendicular,

for gravitaional force, F= mg=5*9.8 N and displacement is 4 m

Work done = force * displacement*cos(theta) {theta is the angle between force and displacement}

WD=5*9.8*4*cos(120) = -98 J

b) According to work energy theorem

kinetic energy at A= potential energy change + kinetic energy at B

kinetic energy at A=0.5*m*v2a=0.5*5*8.12=164.025 J

change in potential energy=mgh= mg*s*sin(alpha) =5*9.8*sin(30)= 24.5 {h=s*sin(alpha)}

kinetic energy at B= 0.5*5*vb2

Plugging the values into Work energy theorem,

164.025=24.5+ 2.5*vb2

vb =7.47 m/s

c) impulse momentum theorem is change in momentum is the impulse

change in momentum= m*va - m*vb =5*(8.1-7.47) =3.15 N-s

impulse, J= 3.15 N-s

net force on the body for time t= mgsin(alpha)+ umgcos(alpha)= mg(sin(alpha) + ucos(alpha))=32.138 N

impusle= Force*time

time= impulse/force= 3.15/32.138 = 0.0980 seconds

d) Force and displacement are at right angles hence no work done

this implies that kinetic energy at B is equal to kinectic energy at C

hence velocity at C = v (say)= 7.47 m/s

f) using force balance on C

mv2/r + R = mg

R is the rection force and r is the radius og=f the circular path

R=mg- mv2/r

R= 49.05 - 139.50

magnitude of reaction force will be = 90.45 N

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