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: 59a

Answer

$a_{min}=\dfrac {M-m}{m}g=\dfrac {15-10}{10}g=\dfrac {g}{2}\approx 4,9\dfrac {m}{s^{2}}$

Work Step by Step

Let's assume the tension on the rope is $T $, mass of monkey is $m$ and mass of the object is $M$ So let's write Newtons second law for the monkey: $\sum \overrightarrow {F}=m\overrightarrow {a}\Rightarrow \overrightarrow {T}+m\overrightarrow {g}=m\overrightarrow {a}$ $m\overrightarrow {g}$ and $\overrightarrow {T}$ are in the opposite direction as shown in the figure (the only force pulls money upwards is T) So when we simplify we get $T-mg=ma (1)$ On the other hand in order to raise the object $T\geq Mg$ so the minimum acceleration needed to raise the object will be when $T=Mg (2)$ So from (1) and (2) we get $a_{min}=\dfrac {M-m}{m}g=\dfrac {15-10}{10}g=\dfrac {g}{2}\approx 4,9\dfrac {m}{s^{2}}$
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