Question

What is the efficiency of a machine that lifts a load of 12.0 kg a vertical distance of 5.00

m in 30.0 s after being supplied with 70.0 W of power? Explain the steps you take to get the answer.

Answer 1

Efficency = 28%

Step-by-step explanation:

We have the load has a weight of:

`w =m*g\\w =12*9.81\\w = 117.72 [N]`

Work in physics is defined as the product of force (weight) by distance.

`W=w*d`

where:

W = work [J]

w = weight = 117.72[N]

d = distance = 5 [m]

`W=117.72*5\\W=588.6[J]`

Now power is defined as the relationship of work at a certain time.

`P=W/t\\`

where:

P = power [W]

W = work = 588.6[J]

t = time = 30[s]

`P=588.6/30\\P=19.62[W]`

Now the efficiency of a machine is defined as the power output over the power input to the machine. The power input should always be greater than the power output.

`efficency = (Power_(out))/(Power_(in))\\Efficeny = (19.62)/(70) \\Efficency = 0.28\\Efficency = 28%`