Please help! this is the second time I posted this question. Hanging Bag A 110.0
ID: 1429182 • Letter: P
Question
Please help! this is the second time I posted this question.
Hanging Bag
A 110.0 kg mail bag hangs by a vertical rope 3.30 m long. A bored (but very strong) postal worker then displaces the bag to a position which is horizontally 2.00 m away from the original vertical line of the rope. During the displacement, the postal worker always keeps the rope taut so that the bag actually moves in an arc, and experiences a vertical displacement in addition to the given horizontal displacement.
A. What horizontal force is necessary to hold the bag in the new position?
B. As the bag is moved to this position, how much work is done by the rope?
C. As the bag is moved to this position, how much work is done by gravity?
D. As the bag is moved to this position, how much work is done by the worker?
Explanation / Answer
Here,
mass , m = 110 Kg
L = 3.3 m
horizontal displacement , h = 2 m
final angle of string and vertical
theta = arcsin(2/3.3)
theta = 37.3 degree
A) let the tension in the string is T
for the force needed to applied
tan(theta) = F/mg
F = m * g * tan(theta)
F = 110 * 9.8 * tan(37.3)
F = 821 N
the force needed is 821 N
b)
as the rope is always taut
it will be perpendicular to motion at all points
work done by rope = F *d * cos(90)
work done by rope =0 J
c)
work done by gravity = - m * g * L * (1 - cos(theta))
work done by gravity = -110 * 9.8 * 3.3 * (1 - cos(37.3))
work done by gravity = -727 J
the work done by gravity is -727 J
d)
the work done by person = - the work done by gravity
the work done by person = 727 J
the the work done by person is 727 J