Answer :
1) 14.7 m
2) 18.4 m
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:
[tex]W_i=W_o[/tex]
where
[tex]W_i=F_id_i[/tex] is the work in input, where
[tex]F_i = 100 N[/tex] is the force applied in input
[tex]d_i[/tex] is the distance through which the input force is applied (which corresponds to the length of the ramp)
[tex]W_o=F_o d_o[/tex] is the work in output, where
[tex]F_o=mg = (50)(9.8)=490 N[/tex] is the force in output, which corresponds to the weight of the box (m = 50 kg is the mass of the box and [tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity)
[tex]d_o = 3 m[/tex] is the distance through which the box has been lifted
Solving for [tex]d_i[/tex], we find the length of the ramp:
[tex]F_i d_i = F_o d_o\\d_i = \frac{F_o d_o}{F_i}=\frac{(490)(3)}{100}=14.7 m[/tex]
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:
[tex]0.80W_i = W_o[/tex]
Which means that the output work is only 80% of the work in input.
The equation can be rewritten as
[tex]0.80F_i d_i = F_o d_o[/tex]
And we have
[tex]F_i = 100 N[/tex]
[tex]F_o = 490 N[/tex]
[tex]d_o = 3m[/tex]
Solving for [tex]d_i[/tex], we find the new length of the ramp:
[tex]d_i = \frac{F_o d_o}{0.80 F_i}=\frac{(490)(3)}{0.80(100)}=18.4 m[/tex]