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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, ZeroTheHero Jul 5 '17 at 20:30

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  • $\begingroup$ 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. $\endgroup$ – honeste_vivere Jul 5 '17 at 13:53

If a person pushes on a cart, the cart will accelerate in the same direction as the applied force. In this case, force and displacement are parallel to each other (not perpendicular). If I push a cart to the west, it will move to the west (not to the north).

When you say "pushes the cart by applying perpendicular force", what is the force perpendicular to? It is not clear what you mean here.

  • $\begingroup$ Or otherwise, you consider the case wherein a person moves on lifting a weight over his head.This means that the force is perpendicular to the displacement vector. Then the work done by the weight on the person will be ZERO. Then will the person feel tired? $\endgroup$ – Vishnu Jul 5 '17 at 4:37
  • $\begingroup$ 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)? $\endgroup$ – Bob Jul 5 '17 at 4:39
  • $\begingroup$ I mean not lifting up and down. What I mean is that the person keeps the weight stationary on his head and walk right or left $\endgroup$ – Vishnu Jul 5 '17 at 4:42
  • $\begingroup$ 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. $\endgroup$ – Bob Jul 5 '17 at 4:45
  • $\begingroup$ Please clarify that the work done by the weight on the person will be ZERO or not as it is mentioned in the situation.Moreover, you are telling that the tiredness felt by the person is only because he WALKS on the ground (due to the opposing frictional force). $\endgroup$ – Vishnu Jul 5 '17 at 4:54

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