Topic:machine,a block and tackle of 6 pulleys is used to raise a load of 300N steadily through a height of 30m.if the work done against friction is 2000J. calculate:the work done by effort,the efficiency of the system,the applied effort.
what effort would be needed to lift a load of 100n
I will be happy to critique your thinking on this.
To calculate the work done by the effort, efficiency of the system, and the applied effort, we can use the following formulas:
1. Work done by effort (W_e) = load lifted (W) - work done against friction (W_f)
2. Efficiency (η) = (output work / input work) × 100%
3. Applied effort (F_a) = output work / distance moved by effort (d_e)
Given:
Load lifted (W) = 300 N
Height (h) = 30 m
Work done against friction (W_f) = 2000 J
Number of pulleys (n) = 6
Let's calculate each value step-by-step:
1. Work done by the effort (W_e):
W_e = W - W_f
= 300 N - 2000 J
= 300 N - 2000 J [since work done (W) = force (F) × distance (d)]
2. Efficiency (η):
To calculate efficiency, we need to find both the input work (W_i) and the output work (W_o).
a. Output work (W_o):
W_o = load lifted (W) × distance (d)
= 300 N × 30 m
= 9000 J
b. Input work (W_i) = work done by effort (W_e) + work done against friction (W_f)
= 300 N - 2000 J [Substituting the values]
Now, we can calculate efficiency (η):
η = (W_o / W_i) × 100%
3. Applied effort (F_a):
F_a = W_o / distance moved by effort (d_e) [To find d_e, we need to determine the length of rope moved by the effort.]
For a block and tackle system with 6 pulleys, the length of rope moved by the effort is equal to the height lifted (as the load is divided equally among the pulleys.)
So, d_e = h = 30 m
Now we can calculate the applied effort (F_a):
F_a = W_o / d_e [Substituting the values]
Please provide the values for work done by the effort, so that I can proceed with the calculations.
To calculate the work done by the effort, the efficiency of the system, and the applied effort, we need to use some formulas and equations related to work, efficiency, and mechanical advantage.
1. Work Done by Effort:
The work done by the effort can be calculated using the formula:
Work = Force x Distance
Given:
Force (effort) = ?
Distance = 30 m
Work = ?
We can rearrange the formula to solve for force:
Force = Work / Distance
Substituting the given values:
Force = 2000 J / 30 m
Therefore, the force (effort) is 66.67 N (rounded to two decimal places).
2. Efficiency of the System:
The efficiency of a machine is the ratio of useful work output to the total work input, expressed as a percentage. The formula for efficiency is:
Efficiency = (Useful Work Output / Total Work Input) x 100
Given:
Useful work output = 300 N x 30 m (force x distance)
Total work input = 2000 J (work done against friction)
Efficiency = ?
We can calculate the useful work output:
Useful Work Output = Force (load) x Distance
Useful Work Output = 300 N (load) x 30 m (distance)
Now, we can substitute the values into the efficiency formula:
Efficiency = (300 N x 30 m) / 2000 J x 100
Therefore, the efficiency of the system is 45%.
3. Applied Effort:
The applied effort is the force exerted by the person or machine to lift the load. In this case, it is the force exerted by the machine. The applied effort force can be calculated using the formula:
Applied Effort = Mechanical Advantage x Load
Given:
Mechanical Advantage = 6 (as there are 6 pulleys)
Load = 300 N
Applied Effort = 6 x 300 N
Therefore, the applied effort is 1800 N.
In summary:
- The work done by the effort is approximately 66.67 N.
- The efficiency of the system is 45%.
- The applied effort is 1800 N.