Work Done by Elevator Cable in Sample Problem 7-6

[Soyuz42]

Homework Statement

An elevator cab of mass m=500 kg is descending with speed v_i=4.0 m/s when its supporting cable begins to slip, allowing it to fall with constant acceleration vec{a}=vec{g}/5 (Figure (a)).
The answer to "(a) During a fall through a distance d=12 m, what is the work done on the cab by the gravitational force vec{F_g}?" is 59 kJ. The answer to "(b) During the 12 m fall, what is the work W_T done on the cab by the upward pull of vec{T} of the elevator cable?" is -47 kJ, and the answer is arrived at by using the free body diagram in Fig. (b) to solve for T, and then using that to solve for W_T. using equation (1) below. My question pertains to (b): why do I get a different answer when I use the work-kinetic energy theorem?

Relevant Equations

(1) W=F*d*cos(phi)
(2) K_f = K_i + W (Work-kinetic energy theorem)
(3) W_g = mgd*cos(phi), where phi is the angle between force and displacement.

As stated, part (a) says that the work done by the gravitational force $\vec{F_g}$ is 59 kJ. If $W_T$ is the work done by the elevator cable during the 12 m fall, then using the work-kinetic energy theorem,
\begin{align*}
K_f -K_i &= W_g + W_T\\
\frac12m({v_f}^2 - {v_i}^2) &= 59000 + W_T\\
\frac12m(a\Delta d)=5886&=59000 + W_T\\
W_T &\approx -53 \text{ kJ},
\end{align*}
while the answer quoted in the text is $-47$ kJ. Why is there a discrepancy?

---

Answer to (b) in the textbook

"A key idea here is that we can calculate the work $W_T$ with Eq. 7-7 ($W=Fd\cos\phi$) if we first find an expression for the magnitude $T$ of the cable's pull. A second key idea is that we can find that expression by writing Newton's second law for components along the $y$ axis in Fig. 7-10b ($F_{\text{net},y} = ma_y$). We get $$T-F_g=ma.$$ Solving for $T$, substituting $mg$ for $F_g$, and then substituting the result in Eq. 7-7, we obtain $$W_T=Td\cos\phi = m(a+g)d\cos\phi.$$ Next, substituting $-g/5$ for the (downward) acceleration $a$ and then $180^\circ$ for the angle $\phi$ between the directions of forces $\vec{T}$ and $m\vec{g}$, we find
\begin{align}
W_T&= m\left(-\frac{g}{5} + g\right) d\cos\phi = \frac45mgd\cos\phi\nonumber\\
&=\frac45(500\text{ kg})(9.8 \text{ m/s^2})(12\text{ m})\cos 180^\circ\nonumber\\
&= -4.70\times 10^4 \text{ J} = -47 \text{ kJ}.\tag{Answer}
\end{align}
(The question and figure are from sample problem 7-6, pg. 149 of Fundamentals of Physics 7e, by Halliday et al.)

 

[Soyuz42]

Never mind, the method was fine, but I incorrectly substituted $a\Delta d$, instead of $2a\Delta d$, for ${v_f}^2 - {v_i}^2$. I would save face but I do not know how to delete this thread.

 

[berkeman]

Never mind, the method was fine, but I incorrectly substituted $a\Delta d$, instead of $2a\Delta d$, for ${v_f}^2 - {v_i}^2$. I would save face but I do not know how to delete this thread.

No need to delete -- it's an interesting problem. Glad you figured it out.

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