A block and tacle of 6 pulley is used to raise a load of 300n steadily through a height of 30m if the workdone against friction is 2000j. calculate the workdone by the effort,the applied effort,the efficiency of the system
work accomplished = 300*30
= 9,000 J
applied work = 9,000 + 2,000 = 11,000 Joules
efficiency = 100 * work accomplished/work applied
= (9/11)100 = 81.8 %
force applied = (300/6)/.818
= 61.1 N
assuming that the 6 in your question refers to mechanical advantage
Yes
Explain pls
Yes
To calculate the work done by the effort, we need to know the formula for work done, which is given by:
Work done = Force x Distance x cos(angle)
In this case, the force is the effort applied, and the distance is the height the load is raised. The angle term does not apply in this scenario.
1. Calculate the work done by the effort:
Given:
Load (force) = 300 N
Height = 30 m
Work done by the effort = Load x Height = 300 N x 30 m = 9000 J
So, the work done by the effort is 9000 J.
2. Calculate the applied effort:
The applied effort is the sum of the work done by the effort and the work done against friction.
Applied effort = Work done by effort + Work done against friction
Applied effort = 9000 J + 2000 J = 11000 J
So, the applied effort is 11000 J.
3. Calculate the efficiency of the system:
Efficiency is defined as the ratio of useful work output to the total work input.
Efficiency = (Useful work output / Total work input) x 100%
Useful work output in this case is the work done by the effort, which is 9000 J. Total work input is the applied effort, which is 11000 J.
Efficiency = (9000 J / 11000 J) x 100%
Efficiency = 0.818 x 100%
Efficiency = 81.8%
Therefore, the efficiency of the system is 81.8%.