Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

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.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com