- #1

weilam06

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- 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??

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??