# Work done by Normal Force

1. Jul 18, 2015

### lavankohsa

1. The problem statement, all variables and given/known data
In a children's park, there is a slide which has a total length of 10 m and a height of 8.0 m .
Vertical ladder are provided to reach the top. A boy weighing 200 N climbs up the ladder to the top of the
slide and slides down to the ground. The average friction offered by the slide is three tenth of his weight. Find
(a) the work done by the ladder on the boy as he goes up, (b) the work done by the slide on the boy as he comes down. Neglect any work done by forces inside the body of the boy.

2. Relevant equations

3. The attempt at a solution
The work done by the ladder is equal to the increase in potential energy of the boy. But in my book it says it is zero.

2. Jul 18, 2015

### phinds

Is the ladder involved in his sliding down?

3. Jul 18, 2015

### lavankohsa

No, but ladder is involved in going up then it should do work when boy goes up

4. Jul 18, 2015

### phinds

what part do the boys leg muscles play relative to the amount of work that it takes him to get to the top?

5. Jul 18, 2015

### lavankohsa

6. Jul 18, 2015

### phinds

7. Jul 19, 2015

### lavankohsa

Ladder doesn't move but does ladder do work on the boy when boy is climbing ? That is I am not able to understand.

8. Jul 19, 2015

### haruspex

A force only does work when the object it acts on moves in the direction of the force. The work done is the integral of the force wrt the displacement. In vectors, that's a dot product: $\int \vec F.\vec {ds}$

9. Jul 19, 2015

### lavankohsa

Ladder applies force on the boy and boy moves up and the displacement is in direction of force applied. That means ladder does work on the boy.

10. Jul 19, 2015

### Qwertywerty

The ladder exerts force on the boy ⊥to itself ( normal ) and along itself ( friction ) .

Normal does no work because it is perpendicular to the diplacement of the boy and friction does no work because it is static and causes no diplacement of the boy .

*Please - I just considered that he climbed along the incline .

Last edited: Jul 19, 2015
11. Jul 19, 2015

### ehild

https://en.wikipedia.org/wiki/Work_(physics)

The point of application of the force has to be displaced in order to do work.
The ladder step exerts force on the foot of the man, but the step does not move.

12. Jul 19, 2015

### haruspex

The point of application of the force from the ladder is on the boy's foot. As the boy moves up, the foot and rung remain stationary. The force does not move the foot, so does no work.
It may help to simplify the scenario by considering the boy's feet and legs as massless, concentrating all his mass in his torso. Treat the ladder, legs and torso as separate bodies. The leg muscles generate a force down on the ladde rung and an equal and opposite force up on the torso. Which force moves the thing it acts on?

13. Jul 19, 2015

### haruspex

No, that's not it. The boy stands on rungs, which are horizontal. The normal force from ladder is therefore vertical.

14. Jul 19, 2015

### Qwertywerty

No sorry - If you would read my edited post I had considered that the boy climbed up the incline .

My mistake .

Last edited: Jul 19, 2015
15. Jul 19, 2015

### Qwertywerty

Anyhow , even if he climbed up the rungs , the only difference would be that the normal force would not be causing any diplacement in the vertical direction and so work done would still be zero .

16. Jul 19, 2015

### lavankohsa

OK, suppose I lift a book from ground then normal force would do work because point of application moves up. And in above question if ladder does no work while climbing then Boy has to do work by internal force . am I correct ?

17. Jul 19, 2015

### Qwertywerty

Yes the boy does work . Effectively , muscular energy of the boy is being conterted to his gravitational potential energy .

18. Jul 19, 2015

### lavankohsa

And i
I have one more confusion. Why we don't include this muscular force of the boy in work -energy theorem.

19. Jul 19, 2015

### Qwertywerty

Okay - Work Energy Theorem →

ΔW = ΔKE .

Change in KE ? Zero .
Work done by gravity ? -ve .
Forces doing work ? The body muscles' and gravity .

So what would be the only way to have ΔKE zero ?

I hope this helps .

Last edited: Jul 19, 2015
20. Jul 19, 2015

### haruspex

What makes you think we don't?