Answer:
vi = 3.95 m/s
Explanation:
We can apply the Work-Energy Theorem as follows:
W = ΔE = Ef - Ei
W = - Ff*d
then
Ef - Ei = - Ff*d <em> </em>
If
Ei = Ki + Ui = 0.5*m*vi² + m*g*hi = 0.5*m*vi² + m*g*hi = m*(0.5*vi² + g*hi)
hi = d*Sin 20º = 5.1 m * Sin 20º = 1.7443 m
Ef = Kf + Uf = 0 + 0 = 0
As we know, vf = 0 ⇒ Kf = 0
Uf = 0 since hf = 0
we get
W = ΔE = Ef - Ei = 0 - m*(0.5*vi² + g*hi) ⇒ W = - m*(0.5*vi² + g*hi) <em> (I)</em>
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If
W = - Ff*d = - μ*N*d = - μ*(m*g*Cos 20º)*d = - μ*m*g*Cos 20º*d <em>(II)</em>
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we can say that
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- m*(0.5*vi² + g*hi) = - μ*m*g*Cos 20º*d
⇒ vi = √(2*g*(μ*Cos 20º*d - hi))
⇒ vi = √(2*(9.81 m/s2)*(0.53*Cos 20º*5.1m - 1.7443 m)) = 3.95 m/s
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