Answer:
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
[tex]\delta A[/tex] = la - lo .............1
[tex]\delta A[/tex] = 0.8 - 0.4
[tex]\delta A[/tex] = 0.4 m
and now we get here stretched length that is
[tex]\delta B[/tex] = lb- lo ........2
[tex]\delta B[/tex] = [tex]\sqrt{0.8^2+0.6^2}[/tex] - 0.4
[tex]\delta B[/tex] = 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 × ( [tex]\delta A[/tex] )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 × ( [tex]\delta B[/tex] )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
