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: 59b

Answer

$\left( M-m\right) g=\left( M+m\right) a\Rightarrow a=\dfrac {\left( M-m\right) }{M+m}g=\dfrac {15-10}{15+10}g=\dfrac {g}{5}\approx 1,96\dfrac {m}{s^{2}}$

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 $M$ When monkey stops object which is heavier will move down and make monkey move up. Let's write Newtons second law for monkey and object For monkey (direction of acceleration will be upwards) $T-mg=ma\left( 1\right) $ For object (acceleration will be down because it is heavier than monkey) $Mg-T=Ma(2)$ The monkey and the object will move with the same magnitude of acceleration because length of rope doesn't change So from (1) and(2) we get $\left( M-m\right) g=\left( M+m\right) a\Rightarrow a=\dfrac {\left( M-m\right) }{M+m}g=\dfrac {15-10}{15+10}g=\dfrac {g}{5}\approx 1,96\dfrac {m}{s^{2}}$
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