1These are basics, guys.. need your help.
A block of mass 'm' is attached to a spring of constant k= 4mg/L and
natural length 10L. The system is released from rest, L above the ground.
Q)
Till the block reaches its lowest position for the first time, the time duration for which the spring remain compressed is- ?
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12 Answers
1Q2)
A tank contains a liquid of density Ï.
A cylinder of density Ï/4 and length L is is kept in equilibrium position by means of an external force, vertically downwards.
The cylinder is just submerged. AT t=0, the external force is removed instantaneously. Assume the liquid level remains same.
The speed of cylinder when it reaches equilibrium position is- ?
Edit: Area of cylinder is 'a' and that of tank is 'A' (a<
49instantaneously means how much time??? I was stuck in a HCV prob due to this instantaneous word!![2][2]
49F=-kx
a=-kx/m=-(4g/L)x
so, a=\frac{dv}{dt}=\frac{dv\: dx}{dx\: dt}=v\frac{dv}{dx}=-\frac{4g}{l}x
\Rightarrow \int_{v_{o}}^{v}{(v)}\:{dv}\;=\;\int_{x_o}^{x} (-\;\frac{4g}{l}x)\: dx
v^{2}-v_o^2=-\frac{4g}{l}(x^2-x_o^2)
which boils down to,
v^{2}=-\frac{4g}{l}x^2+C\;\;\;\;[where\;C=const=v_o^2+\frac{4g}{l}x_o^2]
(\frac{dx}{dt})^{2}=-\frac{4g}{l}x^2+C
this can be used to find the relation between x and t using techniques of integration...[1][1]
49vo has to be found out using equations of motion and xo is 10L
final x ie maximum compression will be found out using conservation of energy...
initial kinetic energy+loss in potential energy=compression in spring...[1][1]
1Thanx subho..can u post the answer to the above question?
I just want to match the answer. [1]
49does that mean that instantaneously removing force equals at the time T=0 the force is removed and never it is applied back!!
1It means that at 't'=0 there is a force acting on the system..and at 't'= t+Δt the force is not acting.
Δt→0
49ok clear thank you!!!
i cnt access instant chat....so wait i ll have to reconnecct modem...net giving problems...:(:(:(