Answer:
3.258 m/s
Explanation:
k = Spring constant = 263 N/m (Assumed, as it is not given)
x = Displacement of spring = 0.7 m (Assumed, as it is not given)
[tex]\mu[/tex] = Coefficient of friction = 0.4
Energy stored in spring is given by
[tex]U=\dfrac{1}{2}kx^2\\\Rightarrow U=\dfrac{1}{2}\times 263\times 0.7^2\\\Rightarrow U=64.435\ J[/tex]
As the energy in the system is conserved we have
[tex]\dfrac{1}{2}mv^2=U-\mu mgx\\\Rightarrow v=\sqrt{2\dfrac{U-\mu mgx}{m}}\\\Rightarrow v=\sqrt{2\dfrac{64.435-0.4\times 8\times 9.81\times 0.7}{8}}\\\Rightarrow v=3.258\ m/s[/tex]
The speed of the 8 kg block just before collision is 3.258 m/s
The speed of the 8kg block before collision on the given rough surface is 2.5 m/s.
The given parameters;
let the spring constant, K = 450 N
let the distance between the blocks, x = 0.5 m
The energy stored in the spring is calculated as follows;
[tex]U = \frac{1}{2} kx^2\\\\U = \frac{1}{2} \times 450 \times 0.5^2\\\\U = 56.25 \ J[/tex]
Apply the principle of conservation of linear momentum, to determine the speed of the 8kg block before collision.
[tex]\frac{1}{2} mv^2 = U - \mu F_n \\\\\frac{1}{2}\times 8 \times v^2 = 56.25 \ - \ 0.4 \times 8\times 9.8\\\\4v^2 = 24.89\\\\v^2 = \frac{24.89}{4} \\\\v^2 = 6.23\\\\v = \sqrt{6.23} \\\\v =2.5 \ m/s[/tex]
Thus, the speed of the 8kg block before collision on the given rough surface is 2.5 m/s.
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