You take a 10 N block and lift it 1 meter vertically from the floor and then place it back down on the floor. How much work do you do during this motion?

W = F x d x cos(theta)
So lifting it does work and the lowering it does negative work won't that cancel out to get OJ of work

Don't know how much work "I" do during this motion because I don't have the equations to calculations human work done during physical activities.

However, work done on the 10N block is zero, the way you calculated it.
When you lift the block, you are doing positive work because θ = 0°, so cos(θ)=1. When you lower the block back to the floor, force is still upwards (against gravity) and displacement is negative (cos(θ)=-1), so work is negative. All in all, there is zero work done (on the block).
The other way to look at it is that the displacement (xf-xi)=0, so work done
=F.D=0 (both are vectors, but scalar multiplication)

To calculate the work done, you can use the formula W = F x d x cos(theta), where W represents the work done, F is the applied force, d is the displacement, and theta is the angle between the force and displacement vectors.

In this case, the force applied is the weight of the block, which is equal to its mass multiplied by the acceleration due to gravity. Assuming the mass of the block is 1 kg (since weight = mass x acceleration due to gravity = 1 kg x 10 m/s^2 = 10 N), the force applied is 10 N.

The displacement is the vertical distance the block is lifted, which is 1 meter.

Furthermore, the angle between the force and displacement vectors is 0 degrees since the force applied is in the same direction as the displacement.

Plugging these values into the formula, we get:

W = 10 N x 1 m x cos(0°)
W = 10 N x 1 m x 1
W = 10 Joules

Therefore, the work done in lifting the block and then placing it back down on the floor is 10 Joules. The negative work done during the lowering does not cancel out the positive work done during the lifting, as they are two separate actions.