A block and tackle with a velocity ratio of 7 is used to raise a mass of 50kg through a vertical distance of 800mm at a steady rate. If the effort is equal to 110N. Determine
(a) The distance moved by the effort.
(b) The workdone by the effort in lifting the load.
(c) The loss in energy involved in operating the machine.
e = L/EVr = 50*9.8/(110*7) = 0.64 = efficiency.
a. W = L*d = (50*9.8)*0.8 = 392 J. = Work done.
e = 392/110d= 0.64
70.4d = 392
d = 5.56 m.
b. 110*5.56 = ___ Joules.
c. (110*5.56)-392 = ___ Joules.
To answer these questions, we need to understand the concepts of velocity ratio, work, and energy.
(a) The distance moved by the effort:
The velocity ratio of a block and tackle system is the ratio of the distance the effort moves to the distance the load moves. In this case, the velocity ratio is given as 7. So, for every 7 units of distance the load moves, the effort moves 1 unit of distance.
We are given that the load (mass) is raised through a vertical distance of 800mm. Let's call the distance moved by the effort as 'x'. According to the velocity ratio, we can set up an equation:
Distance moved by the load / Distance moved by the effort = Velocity Ratio
800mm / x = 7
Solving for x, we get:
x = 800mm / 7
Therefore, the distance moved by the effort is approximately 114.29 mm (rounded to two decimal places).
(b) The work done by the effort in lifting the load:
Work is defined as the force applied over a distance. The work done by the effort is given by the formula:
Work = Force x Distance
In this case, the force (effort) is equal to 110N, and we know the distance moved by the effort is approximately 114.29 mm (converted to meters, it is 0.11429 m). Plugging in these values, we can calculate the work done by the effort:
Work = 110N x 0.11429m
Therefore, the work done by the effort in lifting the load is approximately 12.57 Joules (rounded to two decimal places).
(c) The loss in energy involved in operating the machine:
To determine the loss in energy involved in operating the machine, we need to consider the efficiency of the system. Efficiency is defined as the ratio of useful work output to the total work input. It is usually given as a percentage.
In this case, we are not given the efficiency or any additional information about losses, so we will assume the system is ideal and has 100% efficiency. In an ideal system, there are no losses, so the work input equals the work output.
Therefore, the loss in energy involved in operating the machine is zero.
Note: In real-world applications, machines are not 100% efficient, and energy losses occur due to factors such as friction and heat. However, since we are not provided with any explicit information about losses in this problem, we assume an ideal scenario.