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: 41a

Answer

Magnitude of bundle's acceleration $=1.4$ $m/s^{2}$

Work Step by Step

Weight of the material $=449N=mg$ Rearranging this expression and solving gives: $m=\frac{449}{g} = \frac{449}{9.8}$ $m=45.8$ $kg$ Therefore the mass of the material is $45.8$ $kg$. Let the downward acceleration be $a$. This is the net acceleration of the body and implies that there is a net downward force acting on the body whose magnitude is given by : $F=ma$ And this net force is the result of all the forces in the vertical direction which in this case are the weight of the body in the downward direction and the tension of rope (let it be T) in the upward direction. Therefore balancing the forces and writing the force equation of the system gives: $F = mg - T$ $ma= mg-T$ Substituting known values and setting $T=387 N$ and solving the equation gives: $45.8a = 449-387 $ $a = \frac{62}{45.8} = 1.35$ $m/s^{2}$ $a \approx 1.4$ $m/s^{2}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.