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Question Number 23865 by Tinkutara last updated on 08/Nov/17
A man of mass m, standing at the  bottom of the staircase of height L  climbs it and stands at its top.  (1) Work done by all forces on man is  equal to the rise in potential energy  mgL.  (2) Work done by all forces on man is  zero.  (3) Work done by the gravitational  force on man is mgL.  (4) The reaction force from a step does  not do work because the point of  application of the force does not move  while the force exists.
$$\mathrm{A}\:\mathrm{man}\:\mathrm{of}\:\mathrm{mass}\:{m},\:\mathrm{standing}\:\mathrm{at}\:\mathrm{the} \\ $$$$\mathrm{bottom}\:\mathrm{of}\:\mathrm{the}\:\mathrm{staircase}\:\mathrm{of}\:\mathrm{height}\:{L} \\ $$$$\mathrm{climbs}\:\mathrm{it}\:\mathrm{and}\:\mathrm{stands}\:\mathrm{at}\:\mathrm{its}\:\mathrm{top}. \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Work}\:\mathrm{done}\:\mathrm{by}\:\mathrm{all}\:\mathrm{forces}\:\mathrm{on}\:\mathrm{man}\:\mathrm{is} \\ $$$$\mathrm{equal}\:\mathrm{to}\:\mathrm{the}\:\mathrm{rise}\:\mathrm{in}\:\mathrm{potential}\:\mathrm{energy} \\ $$$${mgL}. \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Work}\:\mathrm{done}\:\mathrm{by}\:{all}\:\mathrm{forces}\:\mathrm{on}\:\mathrm{man}\:\mathrm{is} \\ $$$$\mathrm{zero}. \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Work}\:\mathrm{done}\:\mathrm{by}\:\mathrm{the}\:\mathrm{gravitational} \\ $$$$\mathrm{force}\:\mathrm{on}\:\mathrm{man}\:\mathrm{is}\:{mgL}. \\ $$$$\left(\mathrm{4}\right)\:\mathrm{The}\:\mathrm{reaction}\:\mathrm{force}\:\mathrm{from}\:\mathrm{a}\:\mathrm{step}\:\mathrm{does} \\ $$$$\mathrm{not}\:\mathrm{do}\:\mathrm{work}\:\mathrm{because}\:\mathrm{the}\:\mathrm{point}\:\mathrm{of} \\ $$$$\mathrm{application}\:\mathrm{of}\:\mathrm{the}\:\mathrm{force}\:\mathrm{does}\:\mathrm{not}\:\mathrm{move} \\ $$$$\mathrm{while}\:\mathrm{the}\:\mathrm{force}\:\mathrm{exists}. \\ $$
Commented by Tinkutara last updated on 09/Nov/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$
Commented by ajfour last updated on 08/Nov/17
(2) and (4) , I believe.
$$\left(\mathrm{2}\right)\:{and}\:\left(\mathrm{4}\right)\:,\:{I}\:{believe}. \\ $$
Commented by Tinkutara last updated on 08/Nov/17
Yes! But why?
$$\mathrm{Yes}!\:\mathrm{But}\:\mathrm{why}? \\ $$
Commented by ajfour last updated on 08/Nov/17
let us choose to include in our  system boundary- just the man:  the stairs do no work since the  point of application of force is  there to receive the force and  doesn′t move  as long as the force  is acting there (while the centre    of mass of man moves).  Force of gravity does negative  work since displacement of  centre of mass of man is upwards  while the force of gravity (weight)  is downwards. So (W_g = −mgL).  As our system that includes  only the man has  no change in  its kinetic energy, so work done  by all forces on man is zero.  Work done by forces (muscular)  of man is positive. It leads to a  decrease in his store of internal  energy .  △U_(internal) +△K_(man) =W_g +W_(stairs)   △U_(internal) +0 = −mgL+0  Also △U_(internal) = −W_(man′s muscular forces)  .
$${let}\:{us}\:{choose}\:{to}\:{include}\:{in}\:{our} \\ $$$${system}\:{boundary}-\:{just}\:{the}\:{man}: \\ $$$${the}\:{stairs}\:{do}\:{no}\:{work}\:{since}\:{the} \\ $$$${point}\:{of}\:{application}\:{of}\:{force}\:{is} \\ $$$${there}\:{to}\:{receive}\:{the}\:{force}\:{and} \\ $$$${doesn}'{t}\:{move}\:\:{as}\:{long}\:{as}\:{the}\:{force} \\ $$$${is}\:{acting}\:{there}\:\left({while}\:{the}\:{centre}\:\right. \\ $$$$\left.\:{of}\:{mass}\:{of}\:{man}\:{moves}\right). \\ $$$${Force}\:{of}\:{gravity}\:{does}\:{negative} \\ $$$${work}\:{since}\:{displacement}\:{of} \\ $$$${centre}\:{of}\:{mass}\:{of}\:{man}\:{is}\:{upwards} \\ $$$${while}\:{the}\:{force}\:{of}\:{gravity}\:\left({weight}\right) \\ $$$${is}\:{downwards}.\:{So}\:\left({W}_{{g}} =\:−{mgL}\right). \\ $$$${As}\:{our}\:{system}\:{that}\:{includes} \\ $$$${only}\:{the}\:{man}\:{has}\:\:{no}\:{change}\:{in} \\ $$$${its}\:{kinetic}\:{energy},\:{so}\:{work}\:{done} \\ $$$${by}\:{all}\:{forces}\:{on}\:{man}\:{is}\:{zero}. \\ $$$${Work}\:{done}\:{by}\:{forces}\:\left({muscular}\right) \\ $$$${of}\:{man}\:{is}\:{positive}.\:{It}\:{leads}\:{to}\:{a} \\ $$$${decrease}\:{in}\:{his}\:{store}\:{of}\:{internal} \\ $$$${energy}\:. \\ $$$$\bigtriangleup{U}_{{internal}} +\bigtriangleup{K}_{{man}} ={W}_{{g}} +{W}_{{stairs}} \\ $$$$\bigtriangleup{U}_{{internal}} +\mathrm{0}\:=\:−{mgL}+\mathrm{0} \\ $$$${Also}\:\bigtriangleup{U}_{{internal}} =\:−{W}_{{man}'{s}\:{muscular}\:{forces}} \:. \\ $$
Commented by Tinkutara last updated on 08/Nov/17
So point of application must move  along with COM of man to do some  work?
$$\mathrm{So}\:\mathrm{point}\:\mathrm{of}\:\mathrm{application}\:\mathrm{must}\:\mathrm{move} \\ $$$$\mathrm{along}\:\mathrm{with}\:\mathrm{COM}\:\mathrm{of}\:\mathrm{man}\:\mathrm{to}\:\mathrm{do}\:\mathrm{some} \\ $$$$\mathrm{work}? \\ $$
Commented by ajfour last updated on 08/Nov/17
yes strictly work done by a force  is =∫ F^� .ds^�      where ds^�  is the little  displacement that occurs of the  point where the force F^�  acted  and remained F^�  as long  as   displacement ds^�   took place.
$${yes}\:{strictly}\:{work}\:{done}\:{by}\:{a}\:{force} \\ $$$${is}\:=\int\:\bar {{F}}.{d}\bar {{s}}\:\:\:\:\:{where}\:{d}\bar {{s}}\:{is}\:{the}\:{little} \\ $$$${displacement}\:{that}\:{occurs}\:{of}\:{the} \\ $$$${point}\:{where}\:{the}\:{force}\:\bar {{F}}\:{acted} \\ $$$${and}\:{remained}\:\bar {{F}}\:{as}\:{long}\:\:{as}\: \\ $$$${displacement}\:{d}\bar {{s}}\:\:{took}\:{place}. \\ $$
Commented by Physics lover last updated on 09/Nov/17
Are we assuming that friction  is absent?
$${Are}\:{we}\:{assuming}\:{that}\:{friction} \\ $$$${is}\:{absent}? \\ $$

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