Case 1
If I apply a straight upward(perpendicular to ground) force against gravity of 5N and lift an object "A" 10 meters, then work done is:
$$ W = F \times S = 5\,N \times 10\,m = 50\,J(N-m)$$
Case 2
But if I apply the same amount of force diagonally to the ground, and again push the object "A" 10 meters in the direction of force, then again:
$$ W = F \times S = 5\,N \times 10\,m = 50\,J(N-m) $$
If I apply a straight upward(perpendicular to ground) force against gravity of $5\ \mathrm{N}$ and lift an object "A" 10 meters, then the work done is:
$$ W = F \times S = 5\ \mathrm{N} \times 10\ \mathrm{m} = 50\ \mathrm{J}$$
But if I apply the same amount of force diagonally to the ground, and again push the object "A" 10 meters in the direction of force, then again:
$$ W = F \times S = 5\ \mathrm{N} \times 10\ \mathrm{m} = 50\ \mathrm{J} $$
In case No.cases 1 and 2, the work done is equal, but thethe height would be different in both cases because in case one, height would beis equal to displacement, but in case 2 it will be less than 10 mmeters obviously. Then potential energy (i.e mgh) of both objects would be different, isn't it counter intuitive to. Doesn't conflict with the equation work = energy?
P.S Work is force times displacement ($W = F S \cos(x)$$W = F S \cos(\theta)$); where $x$$\theta$ is angle between direction of force and displacement) and is path independent, according to most of the text books I've read so far.