She will decrease the amount of work
Without the ramp, Ingrid would have to, in no particular order,
• push the box along the ground, which amounts of (20 N) (5 m) = 100 J of work
• lift the box up some height <em>h</em> with a force at least equal in magnitude to the weight of the box <em>mg</em>, so that the work done in the lifting is <em>mgh</em> (i.e. the energy required to lift the box with constant speed to a height <em>h</em>)
and the total work done in this way would be (100 + <em>mgh</em>) J.
With the ramp, Ingrid would instead push the box a length equal to the hypotenuse of the right triangle formed by the two displacements above. Recall that the hypotenuse of any right triangle is always shorter than the sum of its legs' lengths. (This eliminates the third and fourth options.) So if the length of the ramp, or total displacement of the box along the ramp, is <em>s</em>, then the work done by pushing up the ramp is (10 N) <em>s</em> = 10<em>s</em> J
