106(1)
F = -kx2/2
==> a = -kx2/2m
now, vdv = adx
==> ∫vdv = ∫adx
==> v2/2 (limits from 0 to v) = -k/6m*[x3] (limits from 1 to 0.5)
put the values of k and m and get the answer for (a)
1) A particle of mass 10^-2 kg is moving along the positive X - axis under the influence of a force F(x) = -k/2(x^2) where k =10^-2 Nm^2 . At time t=0,it is at x = 1.0 m and its velocity is v=0
a)find its velocity when it reaches x = 0.5 m
b)find the time at which it reaches x = 0.25m
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4 Answers
106
62F(x) = -k/2(x^2)
F=ma(x)
a=-k(x^2)/2m
Till here we have used newton's law
now
a=dv/dt = dv/dt x dx/dx = dv/dx x dx/dt (This is the critical step which is used a lot of times)
Thus, a= v.dv/dx
so
v. dv/dx = -kx2/2m
now follow what asish has done
62hence,
v2/2= - kx3/6m +c
at t=0, x=1 and v=0
so c = k/6m
now you know v in terms fo x
v=dx/dt
now try to solve it!
1for the first part itself you would have got
v=f(x)
now replace v by dx/dt
and integrate to find a relation between x and t
