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.

Answer 1

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)

Answer 2

16

Answer 3

this is a tough question. I do not know the ans

Answer 4

950 plus 50?

Answer 5

Please,the workings isn't clear enough

Answer 6

How did you got MGH=25*9.81*0.40
Especially that 9.81

Answer 7

I don't know the answer go and ask your teacher or your mom or dad. Don't ask me

Answer 8

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

Answer 9

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.