A block of tackle system of pulley consisting of 5 pulleys is used to raise a mass of 25kg through a vertical distance of 40cm at a steady rate. If the effort is equal to 60N, determine:
a) The distance moved by the effort.
b) The work done by the effort in lifting the load.
c) The loss in energy involved in operating the machine.
a) effort distance is five times load distance
b) effort distance multiplied by effort (60N)
c) lost energy = effort work - load work
... load work = m * g * h
To solve this problem, we need to understand the concepts of work, mechanical advantage, and energy. Let's go step by step to find the answers to each question.
a) The distance moved by the effort:
In a block and tackle system, each pulley adds a mechanical advantage. Since the system consists of 5 pulleys, the mechanical advantage is 5. The mechanical advantage represents the ratio of the load force to the effort force. In this case, the load force is the weight of the mass being lifted (25 kg) multiplied by the acceleration due to gravity (9.8 m/s^2).
Load force = mass * acceleration due to gravity
Load force = 25 kg * 9.8 m/s^2
Load force = 245 N
Since the mechanical advantage is 5, the effort force is one-fifth of the load force:
Effort force = Load force / Mechanical advantage
Effort force = 245 N / 5
Effort force = 49 N
To find the distance moved by the effort, we can use the formula:
Mechanical advantage = Distance moved by the effort / Distance moved by the load
Since the mechanical advantage is 5 and the distance moved by the load is given as 40 cm (or 0.4 m), we can rearrange the formula:
Distance moved by the effort = Mechanical advantage * Distance moved by the load
Distance moved by the effort = 5 * 0.4 m
Distance moved by the effort = 2 m
Therefore, the distance moved by the effort is 2 meters.
b) The work done by the effort in lifting the load:
Work is calculated by multiplying the force applied by the distance over which the force is applied. In this case, the effort force is 60 N, and we found in part a) that the distance moved by the effort is 2 m.
Work done by the effort = Effort force * Distance moved by the effort
Work done by the effort = 60 N * 2 m
Work done by the effort = 120 Joules (J)
Therefore, the work done by the effort in lifting the load is 120 Joules.
c) The loss in energy involved in operating the machine:
In an ideal system, there would be no energy loss. However, in real systems, there are energy losses due to factors like friction, heat, and mechanical inefficiencies. In this case, the loss in energy is not explicitly given.
To calculate the loss in energy involved in operating the machine, you would need specific information on the efficiency or any other loss factors involved in the system. Without that information, it's not possible to determine the loss in energy.
Note: In this explanation, I assumed the system is ideal and there are no other variables to consider. The calculations will vary if there are other factors involved.