# Work done by the weight when an object moves down

From University Physics with Modern Physics by Young and Freedman, they say:

We want to find the work done by the weight when the object moves downwards from a height $$y_1$$ above the origin to a lower height $$y_2$$. The weight and displacement are in the same direction, so the work $$W_{\text{grav}}$$ done on the object by its weight is positive: $$W_{\text{grav}} = Fs = w(y_1 - y_2) = mgy_1 - mgy_2$$

This expression also gives the correct work when the object moves upward and $$y_2$$ is greater than $$y_1$$. In that case the quantity $$(y_1 - y_2)$$ is negative, and $$W_{\text{grav}}$$ is negative because the weight and displacement are opposite in direction.

I am confused about this because they have taken displacement, $$s$$, as $$y_1 - y_2$$. I've always learnt that displacement $$s = y_2 - y_1$$ i.e. Final minus initial, not initial minus final.

Is this because they have taken downwards direction as positive? And because down is positive, $$w$$ is positive, and to make $$s$$ positive you do $$y_1 - y_2$$ not $$y_2 - y_1$$? Am I correct? I've been thinking about this, and this interpretation seems to make sense, but what they've done $$s = y_1 - y_2$$ is bothering me because for a very long time I've learnt that displacement is final minus initial never the other way around.

Could you also do this?

For the same scenario, take up as positive, then $$F = -w = -mg$$ and $$W = Fs = -mg(y_2-y_1)$$?

Thank you...

• They have probably used $\vec g = -g$ without telling you. Then they themselves forgot about it and wrote down w(y_1-y_2) instead of $w(y_2-y_1)=-mg(y_2-y_1)= mgy_1-mgy_2$ May 30, 2021 at 15:56