Question

An elevator cage is going up with a velocity of 6 m/s. When the cage was 36 m above the bottom of the shaft, a bolt gets detached from the bottom of the cage floor. Find the velocity with which the bolt strikes the bottom of the shaft and the time that elapses.

Answer 1

Explanation:

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Answer 2

The velocity with which the bolt strikes the bottom of the shaft is 27. 49 m/s.

The time that elapses is 3.35 sec.

Explanation:

Given:

An elevator cage velocity of 6 m/s.

The cage was 36 m above the bottom of the shaft.

To Find:

The velocity with which the bolt strikes the bottom of the shaft.

The time that elapses.

Solution:

As given,an elevator cage velocity of 6 m/s.

An elevator cage velocity u= 6 m/s.

As given,The cage was 36 m above the bottom of the shaft.

The distance of the cage above the bottom of the shaft s=36m.

The velocity with which the bolt strikes the bottom of the shaft =v

Acceleration a =g = $10 m/s^{2}$.

Applying formula  

$v^{2} =u^{2} +2as$          

      $=6^{2} +2\times 10\times6$    

$v^{2} =36 +2\times 10\times 36 =756$

  $v=\sqrt{756} = 27.49m/s$      

Thus, the velocity with which the bolt strikes the bottom of the shaft is 27. 49 m/s.    

 For the time that elapses.

The time that elapses = t

The cage is going up and coming down.

Therefore, considering a= - g and s= -36m

Applying formula  

$s=ut+\frac{1}{2}at^{2}$

$-36=6t+\frac{1}{2}\times(-10)\times t^{2}$

$-36=6t-5 t^{2}$

$t^{2}-6t-36=0$±

$t=\frac{-(-6)\±\sqrt{(-6)^{2} -4\times 5\times(-36)} )}{2\times5}$

  $=\frac{6\±\sqrt{36 -4\times 5\times(-36)} )}{10}$

  $=\frac{6\±\sqrt{36 +720} }{10}$

  $=\frac{6\±\sqrt{756} }{10}$

 $=\frac{6\±27.5 }{10}$

 $= 3.35\ sec.$  

                 (Considering positive sign  because  time can not be negative)

Thus, the time that elapses is 3.35 sec.

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