Answer
Speed with which the bundle hits the ground $=4.1$ $m/s$
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 $
$a = \frac{62}{45.8} = 1.35$ $m/s^{2}$
$a \approx 1.4$ $m/s^{2}$
Now we have the acceleration of the bundle as $1.4$ $m/s^{2}$ and for finding the speed just before hitting the ground, we can use the following equation of kinematics :
$v^{2}=v_{o}^{2} + 2ad$
Substituting the known values and setting $v_{o}=0$ and $d=6.1m$, and solving the equation gives:
$v^{2}=0^{2}+2(1.4)(6.1)$
$v^{2}=17.08$
$v=\sqrt {17.08}=4.13$ $m/s$
$v\approx 4.1$ $m/s$
