Question Number 121434 by mathdave last updated on 08/Nov/20
Commented by mathdave last updated on 08/Nov/20
$${a}\:{help}\:{fom}\:{any}\:{body}\:{pls} \\ $$
Commented by Tawa11 last updated on 06/Sep/21
$$\mathrm{great} \\ $$
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](https://www.tinkutara.com/question/Q121436.png)
$${speed}\:{of}\:{block}\:{before}\:{reaching}\:{rough} \\ $$$${surface}\:={V}_{\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{mV}_{\mathrm{1}} ^{\mathrm{2}} +{mgh}=\frac{\mathrm{1}}{\mathrm{2}}{mV}_{\mathrm{0}} ^{\mathrm{2}} \\ $$$${with}\:{h}={height}\:{of}\:{ramp}={l}_{{ramp}} \:\mathrm{sin}\:\theta \\ $$$${V}_{\mathrm{1}} ^{\mathrm{2}} ={V}_{\mathrm{0}} ^{\mathrm{2}} −\mathrm{2}{gl}_{{ramp}} \mathrm{sin}\:\theta \\ $$$$\Rightarrow{V}_{\mathrm{1}} =\sqrt{{V}_{\mathrm{0}} ^{\mathrm{2}} −\mathrm{2}{gl}_{{ramp}} \mathrm{sin}\:\theta} \\ $$$$=\sqrt{\mathrm{20}^{\mathrm{2}} −\mathrm{2}×\mathrm{10}×\mathrm{2}×\mathrm{sin}\:\mathrm{30}°}=\mathrm{2}\sqrt{\mathrm{95}}\:{m}/{s} \\ $$$$=\mathrm{19}.\mathrm{45}\:{m}/{s} \\ $$$$ \\ $$$${speed}\:{of}\:{block}\:{after}\:{leaving}\:{rough} \\ $$$${surface}\:={V}_{\mathrm{2}} ={before}\:{hitting}\:{spring} \\ $$$${kinetic}\:{energy}: \\ $$$${E}_{{k}} =\frac{\mathrm{1}}{\mathrm{2}}{mV}_{\mathrm{2}} ^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{mV}_{\mathrm{1}} ^{\mathrm{2}} −\mu_{{k}} {mgl}_{{rough}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{mV}_{\mathrm{0}} ^{\mathrm{2}} −{mgl}_{{ramp}} \mathrm{sin}\:\theta−\mu_{{k}} {mgl}_{{rough}} \\ $$$$={m}\left[\frac{\mathrm{1}}{\mathrm{2}}{V}_{\mathrm{0}} ^{\mathrm{2}} −{g}\left({l}_{{ramp}} \mathrm{sin}\:\theta+\mu_{{k}} {l}_{{rough}} \right)\right] \\ $$$$=\mathrm{5}\left[\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{20}^{\mathrm{2}} −\mathrm{10}\left(\mathrm{2}×\mathrm{sin}\:\mathrm{30}°+\mathrm{0}.\mathrm{4}×\mathrm{15}\right)\right] \\ $$$$=\mathrm{650}\:{J} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{Kd}^{\mathrm{2}} ={E}_{{k}} \\ $$$$\Rightarrow{K}=\frac{\mathrm{2}×\mathrm{650}}{\mathrm{2}^{\mathrm{2}} }=\mathrm{325}\:{N}/{m} \\ $$
Commented by Lordose last updated on 08/Nov/20
$$\mathrm{Nice}\:\mathrm{sir}\blacktriangle \\ $$
Commented by peter frank last updated on 08/Nov/20
$$\mathrm{thank}\:\mathrm{you} \\ $$