Answer:
102 m upwards.
Explanation:
Just from a qualitative analysis we can tell the mass it needs to go upwards. How much we determine with the fact that the increase will be - in absolute value - equal to the work gravity does on it to go down that same distance.
Fixed that work being 1 kJ, we get
[tex]\vec F \cdot \vec{ \Delta h} = |F| |\Delta h| cos 0 = ( 9.8 N )\Delta h \cdot 1 = 10^3 J\\\Delta h = \frac {10^3}{9,8} m \approx 102 m[/tex]
