a) Calculate the work done on a 1500-kg elevator car by its cable to lift it 40.0 m at constant speed, assuming friction averages 100 N. b) What is the work done on the lift by the gravitational force in this process? c) What is the total work done on the lift?

The given data can be listed below as,

• The mass of an elevator car is,$m_{car}=1500kg$
• The displacement of car is,$\Delta y=40m$
• The friction force is, $F_s=100N$
• The acceleration due to gravity is, $g=9.8m/s^2$

The free body diagram of an elevator car is as follows:

A non-conservative force's work depends on the path the object takes between its ultimate position and initial position. The amount of work done on an object is determined by multiplying the force's magnitude and projecting the displacement in the force's direction.

Formula of work done is as follows:

$W=fd$

where,

The force f times the distance d equals the work W.

Here,

f = force and d = displacement.

Finding the cable's force is the first step.

Noting that the friction force always acts against the direction in which the object is moving.

Newton's second law teaches us that

$\sum F_y=F_{cable}-mg-F_s=m{a}_y$

where, F is the force, m is the mass, g is the gravity and a is the acceleration.

Now, $a_y=0$, Since the velocity is constant.

$F_{cable}=mg+F_s$

We know that the work done by some force is given by
$W=Fd\cos \theta$

so, the work done by the cable force on the car during this process is given by
$$\begin{gathered} W_{cable}=F_{cable}·\Delta y·\cos 0^o \\ {W}_{cable}=(mg+F_s)\Delta y \\ {W}_{cable}=\left[\right(1500kg\times 9.8m/s^2\times \frac{N}{kg·m/s^2})+100N]\times 40 \\ {W}_{cable}=5.92\times 10J \end{gathered}$$

Formula of work done by the gravitational force is as follows:

$$\begin{gathered} W_{cable}=F_{g}d\cos \theta \\ {W}_{cable}=mg\Delta y\cos \theta \end{gathered}$$

since the angle between the gravitational force and the velocity is ${180}^o$.

Hence,

$W_{gravity}=mg\Delta y\cos {180}^o$

The direction of gravitational force is opposite to the direction of displacement of an elevator car.

Noting that $\cos {180}^o=-1$

$W_{gravity}=mg\Delta y$

Plug the given,

$$\begin{gathered} W_{gravity}=-1500kg\times 9.8m/s^2\times 40.0m\cos {180}^o \\ {W}_{gravity}=-5.88\times {10}^{5}J \end{gathered}$$

The total work done on the lift is the work done by the net force exerted on the car during this process.

Formula of total work done on an elevator car is as follows:
Hence,

$W_{total}=F_{net}\theta y$

We know that the net force exerted on the car during this process is zero since the car is moving at a constant speed

So,

$$\begin{gathered} W_{total}=0\times 10 \\ {W}_{total}=0J \end{gathered}$$

Therefore, total work done on an elevator car is zero.