Answer:
a) [tex]v_p=9.35m/s[/tex]
Explanation:
From the question we are told that:
Open cart of mass [tex]M_o=50.0 kg[/tex]
Speed of cart [tex]V=5.00m/s[/tex]
Mass of package [tex]M_p=15.0kg[/tex]
Speed of package at end of chute [tex]V_c=3.00m/s[/tex]
Angle of inclination [tex]\angle =37[/tex]
Distance of chute from bottom of cart [tex]d_x=4.00m[/tex]
a)
Generally the equation for work energy theory is mathematically given by
[tex]\frac{1}{2}mu^2+mgh=\frac{1}{2}mv_p^2[/tex]
Therefore
[tex]\frac{1}{2}u^2+gh=\frac{1}{2}v_p^2[/tex]
[tex]v_p=\sqrt{2(\frac{1}{2}u^2+gh)}[/tex]
[tex]v_p=\sqrt{2(\frac{1}{2}v_c^2+gd_x)}[/tex]
[tex]v_p=\sqrt{2(\frac{1}{2}(3)^2+(9.8)(4))}[/tex]
[tex]v_p=9.35m/s[/tex]
