91Ó°ÊÓ

The work done is the scalar quantity which is given as the product of the force acting on the body and the displacement.

$W=Fd$

Here, F is the force acting on the body and d is the displacement.

(a)

Free body diagram of the elevator

The force equation on the vertical direction is,

$F_a-mg-f=ma$

Here, $F_a$is the force applied, m is the mass of the elevator$\left(m=1500kg\right)$ , g is the acceleration due to gravity $\left(g=9.8\text{m}/{\text{s}}^2\right)$, f is the frictional force$\left(f=200N\right)$ .

The expression for the force supplied is,

$F_a=ma +mg+f$

Putting all known values,

$\begin{array}{rcl}{F}_a& =& \left(1500\text{kg}\right)×\left(0.8\text{m}/{\text{s}}^2\right)+\left(1500\text{kg}\right)×\left(9.8\text{m}/{\text{s}}^2\right)+\left(200\text{N}\right)\\ & =& 16100\text{N}\\ & & \end{array}$

Therefore, the force must be supplied is $16100N$.

(b)

The work done is,

$W=F_{a}d$

Here,$F_a$is the force supplied$\left(16100N\right)$ , and d is the height the elevator is lifted$\left(d=20.0\text{m}\right)$ .

Putting all known values,

$\begin{array}{rcl}W& =& \left(16100\text{N}\right)×\left(20.0\text{m}\right)\\ & =& 3.22×{10}^5\text{J}\\ & & \end{array}$

Therefore, the work done in lifting the elevator is$3.22×{10}^5\text{J}$ .

(c)

The third equation of motion is,

$v_f=\sqrt{v_i^2+2ad}$

Here,$v_f$is the final speed of the elevator,$v_i$is the initial speed of the elevator ($v_i=0$as the elevator begins from rest), a is the acceleration of the elevator$\left(a=0.8\text{m}/{\text{s}}^2\right)$ , and d is the height the elevator is lifted$\left(d=20.0\text{m}\right)$ .

Putting all known values,

$\begin{array}{rcl}{v}_f& =& \sqrt{\left(0^2\right)+2×\left(0.8\text{m}/{\text{s}}^2\right)×\left(20.0\text{m}\right)}\\ & =& 5.66\text{m}/\text{s}\\ & & \end{array}$

Therefore, the required final speed of the elevator is$5.66m/s$ .

(d)

The work done due to frictional force will go for thermal energy.

The work done due to friction is,

$W=fd$

Here, f is the frictional force$\left(f=200N\right)$ , and d is the height the elevator is lifted$\left(d=20.0m\right)$ .

Putting all known values,

$\begin{array}{rcl}W& =& \left(200\text{N}\right)×\left(20.0\text{m}\right)\\ & =& 4.00×{10}^3\text{J}\\ & & \end{array}$

Therefore, the work went into thermal energy$4.00×{10}^3\text{J}$ .