State the meaning of work done.
A lift (elevator) with weight $13.0\,\text{kN}$ is attached by a cable to a motor, as illustrated in Fig. 2.1. The cable lifts the lift up a vertical shaft. When the lift is moving, a constant frictional force of $2.0\,\text{kN}$ acts on it. The way its speed $v$ varies with time $t$ is shown in Fig. 2.2.
Use Fig. 2.2 to find the lift’s acceleration from $t = 0$ to $t = 3.0\,\text{s}$.
Use Fig. 2.2 to find the work done by the motor in lifting the lift from $t = 3.0\,\text{s}$ to $t = 6.0\,\text{s}$.
The motor’s efficiency is $67\%$. At time $t = 2.5\,\text{s}$, the cable tension is $1.6 \times 10^4\,\text{N}$. Find the input power to the motor at this instant.
State and explain whether the increase in gravitational potential energy of the lift from time $t = 0$ to $t = 7.0\,\text{s}$ is less than, the same as, or greater than the work done by the motor. No calculation is needed.
Use Fig. 2.2 to find the lift’s acceleration from $t = 0$ to $t = 3.0\,\text{s}$.
Use Fig. 2.2 to find the work done by the motor in lifting the lift from $t = 3.0\,\text{s}$ to $t = 6.0\,\text{s}$.
The motor’s efficiency is $67\%$. At $t = 2.5\,\text{s}$, the tension in the cable is $1.6 \times 10^{4}\,\text{N}$. Determine the motor’s input power at this time.
State and explain whether the increase in gravitational potential energy of the lift from time $t = 0$ to $t = 7.0\,\text{s}$ is less than, the same as, or greater than the work done by the motor. No calculation is needed.