# Why is the work done by someone pushing a cart not zero? [duplicate]

We know that when force is perpendicular to displacement, then the work done by it is zero. If this is true, then why do the vegetable vendor (who pushes the cart by applying perpendicular force) feels tired?

## marked as duplicate by Yashas, Kyle Kanos, honeste_vivere, Jon Custer, ZeroTheHeroJul 5 '17 at 20:30

• The pushing is parallel to the displacement, is it not? Further, the vendor must overcome the friction in the wheels/bearings, which is a resistive force. – honeste_vivere Jul 5 '17 at 13:53

• If I lift a weight up 1 meter (suppose the weight begins and ends not moving), the net work done is zero. I can see this directly from the work-kinetic energy theorem ($W_{net}=\Delta KE$). Since both my muscles and gravity are acting on the weight, I may say that $W_{net}=W_{me}+W_{grav}$. Is this what you mean? Or do you mean moving a weight left/right (and the force points downward from gravity and upward due to my muscles)? – Bob Jul 5 '17 at 4:39
• The weight is accelerated when I begin to walk to the right. I have done work on the weight getting it to begin moving ($W_{net}=\Delta KE >0$). If I think of the weight as part of my body, in order to keep moving I must do work against friction to keep moving to the right. When I decide to stop, I must push back against the direction that the weight is moving (push to the left). In all these cases, my muscles were doing work against the weight. – Bob Jul 5 '17 at 4:45