Question

a lift of total mass M kg is raised by cables from rest to rest throug a height h.The greaest tention which the cable can bear safely is nMg newton.Find the shortest interval of time in the accent can be made (n>1)

Answer 1

T - Mg = Ma

As Tmax = n Mg

n Mg - Mg = Ma

Or

a = g (n-1)

s = h

As it starts from rest, u = 0

Second equation of motion:

s = ut + ½ a t²

or

h = 0 + ½ g(n-1) t²

or

t² = 2h /[g(n-1)]

or

t = [ 2h / {g (n-1)} ] 1/2

Answer 2

Answer:

The answer is $v_{m}=\sqrt{\left(2 g h \frac{n-1}{n}\right)}$.

Explanation:

Given,

The greatest tension which the cable can bear safely is nMg newton.

To find,

The shortest interval of time in the accent can be made (n>1)

Weight of lift$=\mathrm{mg}$

Maximum tension$=n M g$

therefore Maximum acceleration $=\frac{n M g-M g}{M}$

$=(n-1) g$

and maximum retardation $=g$

The corresponding velocity-time graph for the shortest time will be as follows Hеге $(n-1) g=\frac{v_{m}}{t_{1}}$ ог

$t_{1}=\frac{v_{m}}{(n-1) g}$(1) and

$g=\frac{v_{m}}{t_{2}}$ or

$t_{2}=\frac{v_{m}}{g} \ldots$ (2)

Aгеа under $v-t$ the graph is total displacement $h$.

Hence $h=\frac{1}{2}\left(t_{1}+t_{2}\right) v_{m}$

From (1), (2), and (3) we get,

$v_{m}=\sqrt{\left(2 g h \frac{n-1}{n}\right)}$

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