Question

(a) What force must be supplied by an elevator cable to produce an acceleration of 0.800 m/s² against a 200-N frictional force, if the mass of the loaded elevator is 1500 kg?

(b) How much work is done by the cable in lifting the elevator 20.0 m?

(c) What is the final speed of the elevator if it starts from rest?

(d) How much work went into thermal energy?

Short answer

(a) The force must be supplied is$16100N$ .

(b) The work done in lifting the elevator is$3.22\times {10}^5\text{J}$ .

(c) The final speed of the elevator is$5.66\text{m}/\text{s}$.

(d) The work went into thermal energy $4.00\times {10}^3\text{J}$.

Step-by-step solution

Step 1: Work done

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.

Force supplied

(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,

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

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

Amount of work done in lifting the elevator

(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,

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

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

Final speed of the elevator

(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,

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

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

Amount of work went into thermal energy

(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,

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

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