The speed of the boy and his friend at the bottom of the slope is 16.52 m/s.
Apply the principle of conservation of energy,
E(up) - E(friction) = E(bottom)
mg sin(15) + ¹/₂(M + m)u² - μ(M + m)cos 15 = ¹/₂(M + m)v²
[tex]v = \sqrt{2[\frac{mgd \ sin15 \ + \frac{1}{2}(M + m)u^2 \ -\mu (M + m)g cos\ 15 }{M + m}] }[/tex]
where;
u = √2gh
u = √(2gL sin15)
u = √(2 x 9.8 x 28 x sin 15)
u = 11.92 m/s
[tex]v = \sqrt{2[\frac{(30)(9.8)(70) \ sin15 \ + \frac{1}{2}(30 + 50)(11.92)^2 \ - 0.12 (30 + 50)9.8 cos\ 15 }{30 + 50}] }\\\\v = 16.52 \ m/s[/tex]
Thus, the speed of the boy and his friend at the bottom of the slope is 16.52 m/s.
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