The spring has an unstretched length of 0.4 m and a stiffness of 200 N/m. The 3-kg slider and attached spring are released from rest at A. Calculate the velocity u of the slider as it reaches B in the absence- of friction, 0.8 m ドーーーー0.6 m

velocity of slider B is 1.537 m/s

Explanation:

given data

unstretched length lo = 0.4 m

stiffness k =  200 N/m

mass  m = 3 kg

solution

we get here first stretched length of spring

= la - lo   .............1

 = 0.8 - 0.4

 = 0.4 m

and now we get here stretched length that is

 = lb- lo   ........2

=    - 0.4

= 0.6 m

and now we get datum height is at b is 0

so total energy at a will be

Energy Ea = Ta + Va  ...................... 3

Ea = 0.5 × m × (va)² + mgh + 0.5 × k × ( )²  

Ea = 0.5 × 3 × (0)² + ( 3 ×9.81 × 0.8 ) + 0.5 × 200 × ( 0.4 )²  

Ea = 39.544 J

and

total energy at b will be

Energy Eb = Tb  + Vb  ...................... 4

Eb = 0.5 × m × (vb)² + mgh + 0.5 × k × ( )²  

Eb = 0.5 × 3 × (vb)² + ( 3 ×9.81 × 0 ) + 0.5 × 200 × ( 0.6 )²  

Eb = ( 1.5× (vb)² + 36  ) J  

so by conservation of energy

Ea = Eb   ...............5

39.544 = ( 1.5× (vb)² + 36  )

solve it we get

vb = 1.537 m/s

answer:

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