Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 5 - Force and Motion-I - Problems - Page 119: 44b

Answer

Tension in cord $=89$ $N$

Work Step by Step

Let $m$ be the mass of the lamp and $T$ be the Tension in the cord. Let the upward direction be $positive$ (sign convention) Net force is given by: $F_{net}=ma$ Writing the force equation using proper sign convention gives: $F_{net}=T-mg$ $ma=T-mg$ Substituting the known values and solving gives: $2.4m=89-9.8m$ $9.8m+2.4m=89$ $12.2m=89$ $m=\frac{89}{12.2} = 7.29$ $kg$ $m\approx 7.3$ $kg$ Now, it is said that acceleration upwards is $2.4$ $m/s^{2}$. Writing the force equation using proper sign convention gives: $F_{net}=T-mg$ $ma=T-mg$ Substituting the known values (now keeping unknown $T$) and solving gives: $7.3kg \times 2.4m/s^{2}=T- 7.3kg \times 9.8m/s^{2}$ $71.5+17.5=T$ $T=89N$ Actually, both, deceleration of $2.4$ $m/s^{2}$ downwards and acceleration of $2.4$ $m/s^{2}$ upwards are the same thing. So, We get the same Tension in cord in both cases.
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