Question

A body of mass 400 kg is suspended at a lower end of a light vertical chain and is being pulled up vertically. Initially the body is at rest and the pull on the chain is 6000 g (g is 9.8 m/s^2) The pull gets smaller uniformly at the rate of 360g N per each meter through which the body is raised. What is the velocity of the body when it has been raised 10m?

Answer 1

Answer:31.62 m/s

Explanation:

Given

mass of body $m=400 kg$

Pull on chain is $F_1=6000g N=60 kN$

Pull get smaller at the rate of $F_2=360g N/m$

Net Upward Force $F=6000 g-360 g\times 10=24 kN$

net acceleration $a=\frac{F}{m}$

$a=\frac{24\times 1000}{m}$

$a=\frac{24000}{400}$

$a=60 m/s^2$

but g is acting downward

$a_{net}=a-g=60-10=50 m/s^2$

using $v^2-u^2=2 as$

here initial velocity is zero

$v^2=2\times 50\times 10$

$v=31.62 m/s$

Answer 2

The velocity of the body will be "31.62 m/s".

Given:

Mass of body,

• m = 400 kg

Pull on chain,

• $F_1= 6000 \ g.N$
• $F_2 = 360 \ g.N/m$

The net upward force will be:

→ $F = 6000-360\times 10$

      $= 6000-3600$

      $= 2400$

or,

      $= 24 \ kN$

Now,

The net acceleration will be:

→ $a = \frac{F}{m}$

     $= \frac{24\times 1000}{400}$

     $= 60 \ m/s^2$

But,

"g" is acting downwards then,

→ $a_{net} = a -g$

          $= 60-10$

          $= 50 \ m/s^2$

By using,

→ $v^2-u^2=2as$

          $v^2= 2\times 50\times 10$

               $= 1000$

            $v = \sqrt{1000}$

               $= 31.62 \ m/s$

Thus the above answer is right.

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