Answer:
1) the required horizontal force F is 1095.6 N
2) W = 0 J { work done by rope will be 0 since tension perpendicular }
3) work is done by the worker is 1029.4 J
Explanation:
Given that;
mass of bag m = 125 kg
length of rope = 3.3 m
displacement of bag d = 2.2 m
1) What horizontal force is necessary to hold the bag in the new position?
from the figure below; ( triangle )
SOH CAH TOA
sin = opp / hyp
sin = d /
sin = 2.2/ 3.3
sin = 0.6666
= sin⁻¹ ( 0.6666 )
= 41.81°
Now, tension in the string is resolved into components as illustrated in the image below;
Tsin = F
Tcos = mg
so
Tsin / Tcos = F / mg
sin / cos = F / mg
we know that; tangent = sine/cosine
so
tan = F / mg
F = mg tan
we substitute
Horizontal force F = (125kg)( 9.8 m/s²) tan( 41.81° )
F = 1225 × 0.8944
F = 1095.6 N
Therefore, the required horizontal force F is 1095.6 N
2) As the bag is moved to this position, how much work is done by the rope?
Tension in the rope and displacement of mass are perpendicular,
so, work done will be;
W = Tdcos90°
W = Td × 0
W = 0 J { work done by rope will be 0 since tension perpendicular }
3) As the bag is moved to this position, how much work is done by the worker
from the diagram in the image below;
SOH CAH TOA
cos = adj / hyp
cos = ( - h) /
we substitute
cos = ( - h) / = 1 - h/
cos = 1 - h/
h/ = 1 - cos
h = ( 1 - cos )
now, work done by the worker against gravity will be;
W = mgh = mf( 1 - cos )
W = mf( 1 - cos )
we substitute
W = (125 kg)((9.8 m/s²)(3.3 m)( 1 - cos41.81° )
W = 4042.5 × ( 1 - 0.745359 )
W = 4042.5 × 0.254641
W = 1029.4 J
Therefore, work is done by the worker is 1029.4 J