Answers are 77.7N, 10.19N, and 35.3 N down. The 1.2-kg slider is released from r

Question

Answers are 77.7N, 10.19N, and 35.3 N down.

The 1.2-kg slider is released from rest in position A and slides without friction along the vertical-plane guide shown. Determine Problem 3/141 the speed VB of the slider as it passes position B the maximum deflection Delta of the spring. The 1.2-kg slider of the system of Prob. 3/141 is released from rest in position A and slides without friction along the vertical-plane guide. Determine the normal force exerted by the guide on the slider just before it passes point C, just after it passes point C, just before it passes point E.

Explanation / Answer

a) for B ...

change in potential = mass * g * ( 3 + 1.5 ) = 1.2 * 9.8 * 4.5 = 52.92 J

so change in potential = kinrticv energy
so.. 52.92 = 0.5 * 1.2 * vB^2

so vB = 9.3915 m / sec

b) for spring...
change in potential = m*g*(3) = 1.2 * 9.8 * 3 = 35.28

this should cause deflection in spring..
so.. 0.5 * 24000 * x^2 = 35.28
so deflection x = 0.054222 m = 5.4222 cm


3/142
a) At C ...
height of C from B = 1.5 * ( 1 - cos 30 ) = 0.201 m ...
so potential energy change from A = m*g* ( 4.5 - 0.201 ) = 1.2*9.8*(4.5-0.201) = 50.55624

so.. 0.5 * 1.2 * vC^2 = 50.55624

so vC = 9.179 m / sec

so just befor C ..
force = mg cos 30 + mvC^2 / 1.5 = 1.2 * [ 9.8*cos 30 + 9.179^2 / 1.5 ] = 77.5877 N

b) just after C ... teh block travels in an incline palne ..
so normal reaction = mg cos 30 = 10.1845 N

c) just before E ..

velocity at E = vE

potential energy change from A = 35.28
0.5 * 1.2 * vE^2 = 35.28
so vE = 7.668 m / sec

so... before E .. .
N =- mgcos 30 + mvE ^2 /1.5 = 36.856 N

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