(a) For the elevator of Example 7.9 (Section 7.2), what is the speed of the elevator after it has moved downward 1.00 m from point 1 in Fig. 7.17? (b) When the elevator is 1.00 m below point 1 in Fig. 7.17, what is its acceleration?

The force constant of the spring is $\mathrm{k}=1.06\times {10}^4\mathrm{N}/\mathrm{m}$

The constant friction force by the clamp is f = 17000 N

The mass of the elevator is m = 2000 kg

The compressed length for spring is y = 1 m

The speed of the elevator at point 1 is u = 4 m/s

The principle of the work-energy theorem is used to solve the above problem. The change in kinetic energy of the elevator between point 1 and mid-point is equal to the work done by the elevator for these positions.

Apply the work-energy theorem to calculate the speed of the elevator:

$\frac{1}{2}\mathrm{m}\left({\mathrm{u}}^2-{\mathrm{v}}^2\right)=\frac{1}{2}{\mathrm{ky}}^2+\left(\mathrm{f}-\mathrm{mg}\right)\mathrm{y}$

Here, m is the mass of the elevator and g is the gravitational acceleration.

Substitute all the values in the above equation and we get,

$$\begin{gathered} \frac{1}{2}\left(2000\mathrm{kg}\right)\left({\left(4\mathrm{m}/\mathrm{s}\right)}^2-{\mathrm{v}}^2\right)=\frac{1}{2}\left(1.06\times {10}^4\mathrm{N}/\mathrm{m}\right){\left(1\mathrm{m}\right)}^2 \\ +\left[17000\mathrm{N}-\left(2000\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(1\mathrm{m}\right)\right] \\ \mathrm{v}=2.82\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the speed of the elevator is 2.82 m/s .

The acceleration of the elevator is calculated as:

${\mathrm{v}}^2={\mathrm{u}}^2-2\mathrm{ay}$

Here, is the acceleration of the elevator

Substitute all the values in the above equation, and we get,

$$\begin{gathered} {\left(2.82\mathrm{m}/\mathrm{s}\right)}^2={\left(4\mathrm{m}/\mathrm{s}\right)}^2-2\mathrm{a}\left(1\mathrm{m}\right) \\ \mathrm{a}=4.02\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the acceleration of the elevator is$4.02\mathrm{m}/{\mathrm{s}}^2$ .