System shown in the figure is released from rest.Pulley and spring is massless and friction is absent everywhere.The speed of 5kg block when 2kg block leaves the contact with ground is(take spring constant=k=40N/m ang g=10m/s^2)
(answer that book gives is 2.828m/s)

Question

Ashwit · Answer

as it is given that system released from rest so kx equals 2g x 0.5m then apply law of conservation of energy for 5 kg block

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System shown in the figure is released from rest.Pulley and spring is massless and friction is absent everywhere.The speed of 5kg block when 2kg block leaves the contact with ground is(take spring constant=k=40N/m ang g=10m/s^2) (answer that book gives is 2.828m/s)

System shown in the figure is released from rest.Pulley and spring is massless and friction is absent everywhere.The speed of 5kg block when 2kg block leaves the contact with ground is(take spring constant=k=40N/m ang g=10m/s^2)
(answer that book gives is 2.828m/s)

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System shown in the figure is released from rest.Pulley and spring is massless and friction is absent everywhere.The speed of 5kg block when 2kg block leaves the contact with ground is(take spring constant=k=40N/m ang g=10m/s^2) (answer that book gives is 2.828m/s)

System shown in the figure is released from rest.Pulley and spring is massless and friction is absent everywhere.The speed of 5kg block when 2kg block leaves the contact with ground is(take spring constant=k=40N/m ang g=10m/s^2) (answer that book gives is 2.828m/s)

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System shown in the figure is released from rest.Pulley and spring is massless and friction is absent everywhere.The speed of 5kg block when 2kg block leaves the contact with ground is(take spring constant=k=40N/m ang g=10m/s^2)
(answer that book gives is 2.828m/s)