Question

An $500\,Kg$ elevator cab is descending with a speed of ${v_0} = 4\,m{\sec ^{ - 1}}$ when the winch system that lowers it begins to slip, allowing it to fall with a constant acceleration of $\dfrac{g}{5}\,m{\sec ^{ - 2}}$ . Curing its fall through a distance of $12\,m$ , the total work done on the elevator cab is (take $g = 10\,m{\sec ^{ - 2}}$ ).
A. $60\,kJ$
B. $ - 40\,kJ$
C. $12\,kJ$
D. $100\,kJ$

Answer 1

Hint:In physics, work done by a body is defined as the scalar product between the force applied on the body the distance by which the body displaces, mathematically its written as $W = \vec F.\vec S$ , work done is a scalar quantity. we will use the work energy theorem which states that “work done on a body is equals to the change in kinetic energy between two points” , so we will find the initial and final kinetic energy of elevator cab and thus the difference between two states will be the net work done by elevator cab.

Complete step by step answer:
Let us first find the time for which elevator cab displaces the distance of $S = 12\,m$ with an initial velocity of ${v_0} = 4\,m{\sec ^{ - 1}}$ and with an acceleration of $a = \dfrac{g}{5}\,m{\sec ^{ - 2}}$. Using, newton’s equation of motion we have,
$S = ut + \dfrac{1}{2}a{t^2}$
Put the values of $S = 12\,m$ and $a = \dfrac{g}{5} = 2\,m{\sec ^{ - 2}}$
We have,
$12 = 4t + {t^2}$
$\Rightarrow {t^2} + 4t - 12 = 0$
$\Rightarrow (t - 2)(t + 6) = 0$
On comparing both factors with zero, since time can’t be negative so,
$t = 2\sec $

Now, after $t = 2\sec $ we need to find the velocity of the elevator cab, we can use the newton’s equation of motion as $v = {v_0} + at$ where we have,
${v_0} = 4\,m{\sec ^{ - 1}}$
$\Rightarrow t = 2\sec $
$\Rightarrow a = 2\,m{\sec ^{ - 2}}$ , on putting the values we get,
$\Rightarrow v = {v_0} + at$
$\Rightarrow v = 4 + 4$
$\Rightarrow v = 8\,m{\sec ^{ - 1}} \\ $
Now, for the journey of $12m$ by elevator cab of mass $m = 500\,Kg$ the initial velocity of the cab was ${v_0} = 4\,m{\sec ^{ - 1}}$ and final velocity is $v = 8m{\sec ^{ - 1}}$ hence, by using work-energy theorem we have,
Work Done $ = $ Final Kinetic Energy $ - $ Initial Kinetic energy.
Put, $K.E = \dfrac{1}{2}m{v^2}$ we get,
$W = \dfrac{1}{2}m[{v^2} - {v_0}^2]$
$\Rightarrow W = 250 \times [64 - 16]$
$\Rightarrow W = 12000\,J$
$\therefore W = 12\,kJ$

Hence, the correct option is C.

Note:It should be remembered that the Work energy theorem is applicable for any type of force whether its conservative force or non-conservative force, and this theorem is widely used when more than one force is acting on a body. A conservative force is one whose work done doesn't depend upon the path followed by the body whereas non-conservative forces are one whose work done depends upon the path followed by the body. Remember, $1\,kJ = 1000\,J$ .