Question

O bila cu m=200g este lansata din A cu viteaza intila (A) de 6 m/s . Pe o pistă . Porțiunea AB =3m. Miscarea pe AB decurge cu frecare unde miu =0.1 . Pe portiunea BC, forta de frecare =0 . Inaltimea R =50cm . Se cere:
a) viteaza in punctul B
b) viteza in punctul C
c) inaltimea maxima

Answer 1

$R=50 cm=0,5 m\\m=200 g=0,2 kg\\a) \Delta Ec_{AB}=L_{F_f}+L_N+L_G\\L_{F_f}}=-F_fl=-\mu mgl\\L_N=0\\L_G=0\\ \Delta Ec_{AB}=E_{c_B}-E_{c_A}=\frac{mv_B^2}{2}-\frac{mv_A^2}{2}\\ \Delta Ec_{AB}=L_{F_f}+L_N+L_G<=>\frac{mv_B^2}{2}-\frac{mv_A^2}{2}=-\mu mgl|:m=>v_B=\sqrt{v_A^2-2\mu gl}=\sqrt{6^2-2*0,1*10}=\sqrt{34}=5,83 m/s=>v_B=5,83 m/s$

$b)c)E_{m_B}=E_{m_C}<=>E_{c_B}+E_{p_B}=E_{c_C}+E_{p_C}\\E_{p_B}=0\\E_{c_B}+E_{p_B}=E_{c_C}+E_{p_C}<=>E_{c_B}=E_{p_C}+E_{c_C}<=>\frac{mv_B^2}{2}=\frac{mv_C^2}{2}+mgR|:m=>v_C=\sqrt{v_B^2-2gR}=\sqrt{34-2*10*0,5}=\sqrt{24}=4,89 m/s=>vv_C=4,89 m/s$

$c)E_{m_B}=E_{m_D}<=>E_{c_B}+E_{p_B}=E_{c_D}+E_{p_D}\\E_{c_C}=0\\E_{p_B}=0\\E_{c_D}+E_{p_B}=E_{c_D}+E_{p_D}<=>E_{c_B}=E_{p_D}<=>\frac{mv_B^2}{2}=mgh_{max}|:m=>h_{max}=\frac{v_B^2}{2g}=\frac{34}{20}= 1,7 m=>h_{max}=1,7 m$