Respuesta :
Answer:
[tex]V_f = 8.286 m/s[/tex]
Explanation:
We will use the energy and the conservation of the linerar mometum for answer this question, so:
First, we will calcule the velocity of the sled when gayle dives in it as:
[tex]P_i = P_f[/tex]
it means that the initial momentum is equal to the finale.
so:
[tex]M_gV_g = M_sV_{si}[/tex]
where [tex]M_g[/tex] is the mass of gayle, [tex]V_g[/tex] is the velocity of gayle, [tex]M_s[/tex] is the mass of the sled and gayle and [tex]V_{si}[/tex] is the velocity of the sled after gayle dives in it
[tex](48 kg)(4.3 m/s) = (48 kg+5.45 kg)V_{si}[/tex]
[tex]V_{si} = 3.86 m/s[/tex]
Now, using the conservation of energy, Â we will calculated what is the velocity of the sled just before the brother of gayle hops on her back. That is:
[tex]M_sgh+\frac{1}{2}M_sV_{si}^2 = \frac{1}{2}M_sV_{sf}^2[/tex]
where g is the gravity, [tex]V_{sf}[/tex] is the velocity of the sled just before the brother of gayle hops in her back and h is the altitude descended.
[tex](48 + 5.45)(9.8)(5)+\frac{1}{2}(48+5.45)(3.86)^2 = \frac{1}{2}(48 + 5.45)V_{sf}^2[/tex]
[tex]V_{sf} = 10.36 m/s[/tex]
Now, we will know the velocity of the sled after the brother of gayle hops on her back using the conservation of the linear momentum as:
[tex]M_sV_{sf} = M_bV_b[/tex]
where [tex]M_b[/tex] is the mass of the sled, gayle and now her brother and [tex]V_b[/tex] is the velocity of the sled after the brother of gayle hops on her back.
[tex](48+5.45)(10.36) = (48+5.45+30)V_b[/tex]
[tex]V_b = 6.63 m/s[/tex]
Finally using the consevation of energy we will find the velocity at the bottom of the hill as:
[tex]\frac{1}{2}M_bV_b^2+M_bgh = \frac{1}{2}M_bV_f^2[/tex]
where h = 10 m and [tex]V_f[/tex] is the final velocity, so:
[tex]\frac{1}{2}(48+5.45+30)(6.63)^2+(48+5.45+30)(9.8)(10) = \frac{1}{2}(48+5.45+30)V_f^2[/tex]
[tex]V_f = 8.286 m/s[/tex]
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