Your answer would be C. Hope this helps!!
Answer:
total work is = 52450 J
Explanation:
given data
mass = 5000-lb
density = 10 lb/ft
height = 50 ft
solution
as we will treat here cable and ball are separate
and
here work need to lift cable is
w = (10Δy )(9.8 y ) j
and
now summing all segment of cable
so passing limit Δy to 0
so total work need
=
=
= 2450J
so lifting 5000 lb wrcking 50 m required additional 5000 + 2450
so total work is = 52450 J
Answer:
Torque, 
Explanation:
It is given that,
Force acting on the particle, 
Position of the particle,
We need to find the torque on the particle about the origin. It is equal to the cross product of position and the force. Its formula is given by :
The cross product of vectors is given by :

or

So, the torque on the particle about the origin
. Hence, this is the required solution.