A 50 N collar slides without friction along a smooth and which is kept inclined at 60o to the horizontal.

The spring attached to the collar and the support C.The spring is unstretched when the roller is at A(AC is horizontal).

Determine the value of spring constant k given that the collar has a velocity of 2.5 m/s when it has moved 0.5 m along the rod as shown in the figure.

#### Solution

Given : W=50 N

AB = AC = 0.5 m

To find : Spring constant

Solution :

Mass of collar = `50/g` kg

Let us assume that h = 0 at position 2

POSITION 1 :

x = 0

E_{s1} = `1/2` x k x x_{1}^{2} = 0

h_{1} = 0.5sin60 = 0.433 m

PE1=mgh1=21.65 J

vA = 0 m/s

KE_{1} = 0J

POSITION II :

v_{B} = 2.5 m/s

PE_{2} = mgh = 0 J (because h=0)

KE_{2} = `1/2` π π π£^{2} = 12 π `50/g` π 2.5^{2}= 15.9276 J

In β³ABC

Applying cosine rule

BC^{2} = AB^{2} + AC^{2} -2 X AB X AC X cos(BAC)

= 0.5^{2} + 0.5^{2} – 2 x 0.5 x 0.5 x cos120

= 0.75

BC = 0.866 m

Un-stretched length of the spring = 0.5 m

Extension of spring(x) = 0.866 - 0.5

=0.366 m

E_{s2} = `1/2` x k x_{2}^{2}= 0.067k

APPLYING WORK ENERGY PRINCIPLE

U_{1-2} = KE_{2} - KE_{1}PE_{1} - PE_{2} + ES_{1} - ES_{2} = KE_{2} - KE_{1}

21.6506-0+0-0.067K=15.9276-0

**K = 85.4343 N/mSPRING CONSTANT IS 85.4343 N/m**