The 5-kg collar has a velocity of 5 m>s to the right when it is at A. It then travels along the smooth guide. Determine its speed when its center reaches point B and the normal force it exerts on the rod at this point. The spring has an unstretched length of 100 mm and B is located just before the end of the curved portion of the rod.

Explanation:

mass of collar = 5-kg

velocity at point A = 5 m/s

g = 9.81 ( acceleration due to gravity )

height = 200 mm = 0.2 m

To determine the speed and force at point B

first we get the potential energy at points A and point B

Pa = m*g*h = 5 * 9.81 * 0.2 = 9.8 J

Pb = 0 because velocity is zero at that point.

second we get the velocity at the points when the spring stretches

Xa = = 0.1828

Xb = 0.4 - 0.1 = 0.3

thirdly we calculate the kinetic energy at points A and B

Ka = 1/2 m = =  0.8354 j

Kb = 1/2 m = = 2.25 j

fourthly we calculate for the speed at point B using the conservation energy formula:

Ta + Va = Tb + Vb

+ Pa + Xa = + Pb + Xb

62.5 + 9.9828 = 2.25 +   therefore

Vb = 5.325 m/s

The normal force exerted on the rod at point B

   mass = 5  = 693.95 N

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answer:

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