Question

1. If you want to use a ramp to lift a 50 kg box up 3 m using 100 N of force, how long should the ramp be?

2. If the ramp is only 80% efficient, due to friction, how long would it need to be?

Answer 1

1) 14.7 m

2) 18.4 m

Step-by-step explanation:

1)

Assuming that the ramp is 100%, we can apply the law of conservation of energy: this means that the work done in input to lift the box must be equal to the work done in output.

So we can write:

`W_i=W_o`

where

`W_i=F_id_i` is the work in input, where

`F_i = 100 N` is the force applied in input

`d_i` is the distance through which the input force is applied (which corresponds to the length of the ramp)

`W_o=F_o d_o` is the work in output, where

`F_o=mg = (50)(9.8)=490 N` is the force in output, which corresponds to the weight of the box (m = 50 kg is the mass of the box and
`g=9.8 m/s^2` is the acceleration due to gravity)

`d_o = 3 m` is the distance through which the box has been lifted

Solving for
`d_i`, we find the length of the ramp:

`F_i d_i = F_o d_o\\d_i = (F_o d_o)/(F_i)=((490)(3))/(100)=14.7 m`

2)

In this case, the ramp is only 80% efficient, due to the presence of friction forces. This means that only 80% of the work done in input is converted into work in output.

Therefore in this case, we can write:

`0.80W_i = W_o`

Which means that the output work is only 80% of the work in input.

The equation can be rewritten as

`0.80F_i d_i = F_o d_o`

And we have

`F_i = 100 N`

`F_o = 490 N`

`d_o = 3m`

Solving for
`d_i`, we find the new length of the ramp:

`d_i = (F_o d_o)/(0.80 F_i)=((490)(3))/(0.80(100))=18.4 m`