A block and tackle with a velocity ratio of 5 is used to raise a mass of 25kg through a vertical distances of 40cm at a steady rate.if the effort is equal to 60N,determine.(a)The distance moved by the effort.(b)The workdone by the effort in lifting the load.(c)The loss in energy involve in operating the machine.
If the block moves .40 meters then the line is pulled 5 * .4 = 2 meters (a)
the work done by the puller person on that line is
W in = 60 N * 2 meters = 120 Joules (b)
the useful work done on the block was to lift it .40 meters
work out = m g h = 25 * 9.81 * .4 = 98.1 Joules
so the loss due to friction was
120 - 98.1 = 21.9 Joules (c)
16
this is a tough question. I do not know the ans
Please,the workings isn't clear enough
Velocity ratio (V.R) = 5
Mass (M) = 25 kg
Distance (Dv) =40 cm
Force (F) = 60 N
De = V.R × D.V
DE=(5)×(40)
= 200/100 =2.0 M
W=F×DE
W={60 N} × {200 CM}
(60N) × {(200CM) ( 1M/100CM)}
= 120 JOULES
The distance moved by the effort (De) is 2.0 meters. (a)
The work done by the effort in lifting the load (W) is 120 Joules. (b)
To find the loss in energy involved in operating the machine, we need to calculate the work done against gravity by the effort, which is the useful work done, and subtract it from the work done by the effort.
The useful work done is given by the formula: W = mgh, where m is the mass, g is acceleration due to gravity (9.81 m/s^2), and h is the vertical distance.
W = (25 kg) × (9.81 m/s^2) × (0.40 m) = 98.1 Joules
The loss in energy is then: loss = W - 120 Joules = 98.1 Joules - 120 Joules = -21.9 Joules. (c)
Note: The negative sign indicates a loss in energy.
To solve this problem, we need to understand some key concepts.
First, the velocity ratio of a block and tackle is the ratio of the distance the effort moves to the distance the load moves. In this case, the velocity ratio is given as 5.
The distance moved by the effort refers to the vertical distance covered by the effort in lifting the load.
The work done by the effort in lifting the load can be calculated using the formula:
Work = Force × Distance
where the force is the effort applied, and the distance is the distance moved by the effort.
The loss in energy involved in operating the machine refers to the energy lost due to friction and other inefficiencies in the system.
Now, let's solve the problem step by step:
(a) The distance moved by the effort:
Since the velocity ratio is given as 5 and the load is raised a vertical distance of 40cm, we can calculate the distance moved by the effort as follows:
Distance moved by the effort = Velocity ratio × Vertical distance moved by the load
Distance moved by the effort = 5 × 40cm = 200cm
Therefore, the distance moved by the effort is 200cm.
(b) The work done by the effort in lifting the load:
The work done by the effort can be calculated using the formula:
Work = Force × Distance
Given the effort is equal to 60N and the distance moved by the effort is 200cm, we need to convert the distance to meters before calculating the work done:
Distance moved by the effort = 200cm = 200/100 = 2 meters
Work = 60N × 2m = 120 Joules
Therefore, the work done by the effort in lifting the load is 120 Joules.
(c) The loss in energy involved in operating the machine:
The loss in energy involved in operating the machine refers to the energy lost due to friction and other inefficiencies in the system. In this case, it is not given explicitly. However, we can assume that the efficiency of the machine is 100% (meaning no energy loss), so the loss in energy would be zero.
Therefore, the loss in energy involved in operating the machine is zero.
In summary:
(a) The distance moved by the effort is 200cm.
(b) The work done by the effort in lifting the load is 120 Joules.
(c) The loss in energy involved in operating the machine is zero.
How did you got MGH=25*9.81*0.40
Especially that 9.81