Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 121434 by mathdave last updated on 08/Nov/20

Commented by mathdave last updated on 08/Nov/20

a help fom any body pls

$${a}\:{help}\:{fom}\:{any}\:{body}\:{pls} \\ $$

Commented by Tawa11 last updated on 06/Sep/21

great

$$\mathrm{great} \\ $$

Answered by mr W last updated on 08/Nov/20

speed of block before reaching rough  surface =V_1   (1/2)mV_1 ^2 +mgh=(1/2)mV_0 ^2   with h=height of ramp=l_(ramp)  sin θ  V_1 ^2 =V_0 ^2 −2gl_(ramp) sin θ  ⇒V_1 =(√(V_0 ^2 −2gl_(ramp) sin θ))  =(√(20^2 −2×10×2×sin 30°))=2(√(95)) m/s  =19.45 m/s    speed of block after leaving rough  surface =V_2 =before hitting spring  kinetic energy:  E_k =(1/2)mV_2 ^2 =(1/2)mV_1 ^2 −μ_k mgl_(rough)   =(1/2)mV_0 ^2 −mgl_(ramp) sin θ−μ_k mgl_(rough)   =m[(1/2)V_0 ^2 −g(l_(ramp) sin θ+μ_k l_(rough) )]  =5[(1/2)×20^2 −10(2×sin 30°+0.4×15)]  =650 J    (1/2)Kd^2 =E_k   ⇒K=((2×650)/2^2 )=325 N/m

$${speed}\:{of}\:{block}\:{before}\:{reaching}\:{rough} \\ $$$${surface}\:={V}_{\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{mV}_{\mathrm{1}} ^{\mathrm{2}} +{mgh}=\frac{\mathrm{1}}{\mathrm{2}}{mV}_{\mathrm{0}} ^{\mathrm{2}} \\ $$$${with}\:{h}={height}\:{of}\:{ramp}={l}_{{ramp}} \:\mathrm{sin}\:\theta \\ $$$${V}_{\mathrm{1}} ^{\mathrm{2}} ={V}_{\mathrm{0}} ^{\mathrm{2}} −\mathrm{2}{gl}_{{ramp}} \mathrm{sin}\:\theta \\ $$$$\Rightarrow{V}_{\mathrm{1}} =\sqrt{{V}_{\mathrm{0}} ^{\mathrm{2}} −\mathrm{2}{gl}_{{ramp}} \mathrm{sin}\:\theta} \\ $$$$=\sqrt{\mathrm{20}^{\mathrm{2}} −\mathrm{2}×\mathrm{10}×\mathrm{2}×\mathrm{sin}\:\mathrm{30}°}=\mathrm{2}\sqrt{\mathrm{95}}\:{m}/{s} \\ $$$$=\mathrm{19}.\mathrm{45}\:{m}/{s} \\ $$$$ \\ $$$${speed}\:{of}\:{block}\:{after}\:{leaving}\:{rough} \\ $$$${surface}\:={V}_{\mathrm{2}} ={before}\:{hitting}\:{spring} \\ $$$${kinetic}\:{energy}: \\ $$$${E}_{{k}} =\frac{\mathrm{1}}{\mathrm{2}}{mV}_{\mathrm{2}} ^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{mV}_{\mathrm{1}} ^{\mathrm{2}} −\mu_{{k}} {mgl}_{{rough}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{mV}_{\mathrm{0}} ^{\mathrm{2}} −{mgl}_{{ramp}} \mathrm{sin}\:\theta−\mu_{{k}} {mgl}_{{rough}} \\ $$$$={m}\left[\frac{\mathrm{1}}{\mathrm{2}}{V}_{\mathrm{0}} ^{\mathrm{2}} −{g}\left({l}_{{ramp}} \mathrm{sin}\:\theta+\mu_{{k}} {l}_{{rough}} \right)\right] \\ $$$$=\mathrm{5}\left[\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{20}^{\mathrm{2}} −\mathrm{10}\left(\mathrm{2}×\mathrm{sin}\:\mathrm{30}°+\mathrm{0}.\mathrm{4}×\mathrm{15}\right)\right] \\ $$$$=\mathrm{650}\:{J} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{Kd}^{\mathrm{2}} ={E}_{{k}} \\ $$$$\Rightarrow{K}=\frac{\mathrm{2}×\mathrm{650}}{\mathrm{2}^{\mathrm{2}} }=\mathrm{325}\:{N}/{m} \\ $$

Commented by Lordose last updated on 08/Nov/20

Nice sir▲

$$\mathrm{Nice}\:\mathrm{sir}\blacktriangle \\ $$

Commented by peter frank last updated on 08/Nov/20

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com