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Question-178875




Question Number 178875 by Tawa11 last updated on 22/Oct/22
Commented by mr W last updated on 22/Oct/22
F_(average) =((mv^2 )/(2s))=((0.05×44^2 )/(2×0.04))=1210N  ⇒C
$${F}_{{average}} =\frac{{mv}^{\mathrm{2}} }{\mathrm{2}{s}}=\frac{\mathrm{0}.\mathrm{05}×\mathrm{44}^{\mathrm{2}} }{\mathrm{2}×\mathrm{0}.\mathrm{04}}=\mathrm{1210}{N} \\ $$$$\Rightarrow{C} \\ $$
Commented by Tawa11 last updated on 22/Oct/22
God bless you sir. I understand now.  K.E  =  Workdone.  I appreciate sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{understand}\:\mathrm{now}. \\ $$$$\mathrm{K}.\mathrm{E}\:\:=\:\:\mathrm{Workdone}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$
Commented by mr W last updated on 22/Oct/22
during the strike the contact force  between the club and the golf is not  constant. the work done by this force  is W=∫_0 ^s Fds=F_(average) s. we can only  determine the average force.
$${during}\:{the}\:{strike}\:{the}\:{contact}\:{force} \\ $$$${between}\:{the}\:{club}\:{and}\:{the}\:{golf}\:{is}\:{not} \\ $$$${constant}.\:{the}\:{work}\:{done}\:{by}\:{this}\:{force} \\ $$$${is}\:{W}=\int_{\mathrm{0}} ^{{s}} {Fds}={F}_{{average}} {s}.\:{we}\:{can}\:{only} \\ $$$${determine}\:{the}\:{average}\:{force}. \\ $$
Commented by Tawa11 last updated on 22/Oct/22
God bless you sir. Thanks for your time.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time}. \\ $$

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