## College Physics (4th Edition)

(a) $T = 10.78~m$, (where $m$ is the mass of the truck) Note that $T = 1.10~mg$ (b) $T = 10.78~m$, (where $m$ is the mass of the truck) Note that $T = 1.10~mg$
(a) If the downward speed of the truck is decreasing at a rate of $0.10~g$, then the upward acceleration is $0.10~g$. We can find the tension in the cable: $\sum F = ma$ $T-mg = ma$ $T = m~(a+g)$ $T = m~(0.10~g+g)$ $T = m~(1.10~g)$ $T = m~(1.10)(9.80)$ $T = 10.78~m$, (where $m$ is the mass of the truck) Note that $T = 1.10~mg$ (b) If the downward speed of the truck is decreasing at a rate of $0.10~g$, then the upward acceleration is $0.10~g$. We can find the tension in the cable: $\sum F = ma$ $T-mg = ma$ $T = m~(a+g)$ $T = m~(0.10~g+g)$ $T = m~(1.10~g)$ $T = m~(1.10)(9.80)$ $T = 10.78~m$, (where $m$ is the mass of the truck) Note that $T = 1.10~mg$