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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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