Work done by an electric motor to drive elevator

[weilam06]

Homework Statement

An elevator consists of an elevator cage and a counterweight attached to the ends of a cable that runs over a pulley. The mass of the cage (with its load) is $1200$ kg, and the mass of the counterweight is $1000$ kg. The elevator is driven by an electric motor attached to the pulley. Suppose that the elevator is initially at rest on the first floor of the building and the motor makes the elevator accelerate upward at the rate of $1.5 \text{m/s}^2$.

Relevant Equations

$\sum F=ma$
$x=x_0+v_0t+\frac{1}{2}at^2$
$W=F_x\Delta x$

https://www.physicsforums.com/attachments/250022
For the part (a), set the tension of the string that pulls the elevator be $T_1$ and the tension that pushes the counterweight be $T_2$. Then we have the following equations:

$$T_1−1200g=1200 \cdot 1.5$$
$$1000g−T_2=1000 \cdot (−1.5)$$

where $g=9.8 ms^{−2}$ stands for gravitational acceleration and define the upwards direction be positive #y# value. This gives $T_1=13650 N$ and $T_2=8300 N$. For part (b), since the elevator is driven with net acceleration $a=1.5 ms^{−2}$, so the total distance traveled is
$$d=12 \cdot (1.5+g) \cdot 12=5.65 \text{m}$$
whereas the counterweight traveled $-5.65$ m, with similar approach above. So the work done by the electric motor is
$$W=13650 \times 5.65+8300 \times (−5.65)=302275J$$

which is different from the given answer $4300 \text{J}$. What am I doing wrong here??

 

[haruspex]

so the total distance traveled is

You seem to be using some data not present in your problem statement. If you are given the time of 12 seconds then I don't understand why have a g in there, and there's a factor 1/2 missing.

I tried the link, but it is broken.

 

[weilam06]

Sorry about that. The question asked for
(a) What is the tension in the part of the cable attached to the elevator cage? What is the tension in the part of the cable attached to the counterweight?
(b) The acceleration lasts exactly 1.0 s. How much work has the electric motor done in this interval? Ignore friction forces and ignore the mass of the pulley.
(c) After the acceleration interval of 1.0 s, the motor pulls the elevator upward at constant speed until it reaches the third
floor, exactly 10.0 m above the first floor. What is the total amount of work that the motor has done up to this point?

 

[Delta2]

For b) I believe you should use the work energy theorem
$$\Delta K=W_{T_1}+W_{counterweight}-W_{chamberweight}-W_{T_2}+W_{motor}$$

where $\Delta K=\frac{1}{2}(m_{chamber}+m_{counterweight})v^2-0$ since the system starts from initial velocity=0.

But first you should calculate correctly the distance traveled. You say it is $d=12(1.5+g)12$. Please explain how you arrive at this equation because I don't seem to understand. Isn't the distance traveled simply $d=\frac{1}{2}at^2$ where $a=1.5m/s^2$ and $t=1.0s$?

 

[haruspex]

For b) I believe you should use the work energy theorem
$$\Delta K=W_{T_1}+W_{counterweight}-W_{chamberweight}-W_{T_2}+W_{motor}$$

Why are you adding W_motor? Isn't that already accounted for by the tensions?

 

[haruspex]

1000g−T2=1000⋅(−1.5)​

Think about that. Is T2 more or less than 1000g kg?

the given answer 4300J

For part b I get a bit under 4000J.

 

[weilam06]

But first you should calculate correctly the distance traveled. You say it is $d=\frac{1}{2}(1.5+g)1^2$. Please explain how you arrive at this equation because I don't seem to understand. Isn't the distance traveled simply $d=\frac{1}{2}at^2$ where $a=1.5m/s^2$ and $t=1.0s$?

Because this is due to my frame of reference, and taking account into the gravitational acceleration on Earth. I believed the distance traveled that I calculated has no wrong.

 

[Delta2]

Because this is due to my frame of reference, and taking account into the gravitational acceleration on Earth. I believed the distance traveled that I calculated has no wrong.

Ok, first of all you have typos there, you mean $d=\frac{1}{2}(1.5+g)1^2$. Second, in what frame of reference you need to add g? You take a frame of reference where we free fall? Why to do that??
In the frame of reference of ground, the chamber moves upwards with acceleration 1.5m/s^2 and initial velocity v=0, hence the distance traveled is $d=\frac{1}{2}at^2=0.75m$ there can't be more simple than that.

Why are you adding W_motor? Isn't that already accounted for by the tensions?

Ok i think i see the work of the motor is $W_{motor}=W_{T_1}-W_{T_2}$

 

[haruspex]

Because this is due to my frame of reference,

Work done can be different in different reference frames, even if inertial. You must use the ground frame.
Consider an object mass m accelerated from v₁ to v₂ in one frame. Compare the work done in that frame with that in the inertial frame in which it was initially at rest.

 

[neilparker62]

For part b I get a bit under 4000J.

Same here.