System shown in figure is released from rest with the spring at natural length. Pulley and spring is massless and friction is absent everywhere. The speed $$($$ in $$m/s)$$ of $$5$$ kg block when $$2$$ kg block leaves the contact with ground is (Take$$ force constant of spring $$K=40$$ $$N/m$$ and $$g=10$$ $$m/s2$$,$$2=1.414)

Question

System shown in figure is released from rest with the spring at natural length. Pulley and spring is massless and friction is absent everywhere. The speed $$($$ in $$m/s)$$ of $$5$$ kg block when $$2$$ kg block leaves the contact with ground is (Take$$ force constant of spring $$K=40$$ $$N/m$$ and $$g=10$$ $$m/s2$$,$$2=1.414)

Infinity Learn · Accepted Answer

$$1$$.  $$2$$ kgblock will leave the contact if $$T=mg(m=2$$ $$kg)Also$$, $$T=kx0$$⇒$$x0=12m2$$.     From energy conservation for system Decrease in gravitational potential energy $$=$$ Increase in spring potential energy $$+$$ Increase in kinetic $$energy3$$.$$mgx=12kx2+12mv2where$$ $$m=5$$ kg,$$x=12meter$$,$$k=40N/m$$⇒$$5$$×$$10$$×$$12=12$$×$$40$$×$$(12)2+12$$×$$5v2$$⇒$$40=5v2$$⇒$$v2=8$$⇒$$v=22Therefore$$, the correct answer is $$2.83$$

System shown in figure is released from rest with the spring at natural length. Pulley and spring is massless and friction is absent everywhere. The speed ( in m/s) of 5 kg block when 2 kg block leaves the contact with ground is (Take force constant of spring K=40 N/m and g=10 m/s2,2=1.414)

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