Soyuz42
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- 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.)
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