<h2>The man have to apply force of 160 N</h2>
Explanation:
The work done to lift the bag of weight mg through height 2.5 m is 400 J
The work done can be found by relation W = mg x h
Thus mg =
=
= 160 N
Therefore the man have to apply the force of 160 N
F=MA
F=(8 kg)(9.8 m/s)
F= 78.4 N
W=FD
W=(78.4 N)(7 m)
W=548.8 J
How this helps
Explanation :
Static friction is the frictional force between two objects that are at rest. While sliding friction is the frictional force between two objects in contact and are sliding w.r.t each other.
Static friction is usually greater than sliding friction because in static friction the contact forces is more and the interlocking between objects is tight as compared to sliding friction.
Answer:
The answer of the part (a) is v2 = 7.09 m/s
and the answer of the part (b) is vA1 = 5.25 m/s
Explanation:
Explanation of the both parts of answer is in the following attachments
Answer
given,
mass of the person, m = 50 Kg
length of scaffold = 6 m
mass of scaffold, M= 70 Kg
distance of person standing from one end = 1.5 m
Tension in the vertical rope = ?
now equating all the vertical forces acting in the system.
T₁ + T₂ = m g + M g
T₁ + T₂ = 50 x 9.8 + 70 x 9.8
T₁ + T₂ = 1176...........(1)
system is equilibrium so, the moment along the system will also be zero.
taking moment about rope with tension T₂.
now,
T₁ x 6 - mg x (6-1.5) - M g x 3 = 0
'3 m' is used because the weight of the scaffold pass through center of gravity.
6 T₁ = 50 x 9.8 x 4.5 + 70 x 9.8 x 3
6 T₁ = 4263
T₁ = 710.5 N
from equation (1)
T₂ = 1176 - 710.5
T₂ = 465.5 N
hence, T₁ = 710.5 N and T₂ = 465.5 N