Engineering Mechanics: Statics and Dynamics, 14th Edition
Authors: Russell C. Hibbeler
ISBN-13: 978-0133915426
See our solution for Question 8P from Chapter 14 from Hibbeler's Engineering Mechanics.
Problem 8P
Chapter:
Problem:
A force of F = 250 N is applied to the end at B. Determine the...
Step-by-Step Solution
Step 1
Step 2
Step 3
We are given the mass $m = 10\,{\rm{kg}}$, force $F = 250\,{\rm{N}}$ and distance $d = 1.5\;{\rm{m}}$.
We are asked to determine the final speed of block.
Step 2
The following is the free body diagram:
The distance between pulley is calculated as:
\[\begin{array}{l} l = \frac{d}{2}\\ l = \left( {\frac{{1.5\;{\rm{m}}}}{2}} \right)\\ l = 0.75\;{\rm{m}} \end{array}\]To find the net work done we will use the relation,
\[\Sigma U = Fl - Wd\]Step 3
To calculate the velocity, we will use the wok energy principle.
\[\begin{array}{c} E + \Sigma U = E'\\ \left( {\frac{1}{2}m{v^2}} \right) + \left( {Fl - Wd} \right) = \left( {\frac{1}{2}m{v_0}^2} \right)\\ \left( {\frac{1}{2}m{v^2}} \right) + \left( {Fl - mgd} \right) = \left( {\frac{1}{2}m{v_0}^2} \right) \end{array}\]Here, ${v_0}$ is the required speed of the block and $v$ is the initial speed which is equal to zero.
On plugging the values in the above relation, we get,
\[\begin{array}{c} \left( {\frac{1}{2}\left( {10\;{\rm{kg}}} \right){{\left( {0\;{\rm{m/s}}} \right)}^2}} \right) + \left( {250\,{\rm{N}}} \right)\left( {\frac{{1{\rm{ kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}{{1{\rm{ N}}}}} \right) \times \left( {0.75\,{\rm{m}}} \right) - \left( {10\,{\rm{kg}}} \right) \times \left( {9.81\,{\rm{m/}}{{\rm{s}}^2}} \right) \times \left( {1.5\,{\rm{m}}} \right) = \left( {\frac{1}{2}\left( {10\;{\rm{kg}}} \right){v_0}^2} \right)\\ {v_0} = 2.8\;{\rm{m/s}} \end{array}\]