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A 1 kg block of wood is attached to a spring, of force constant 200 N/m, which is attached to an immovable support. The block rests on a frictional surface with a coefficient of kinetic friction of 0.2. A 20 g bullet is fired into the block horizontally compressing the spring a maximum distance of 15 cm. Find the original velocity of the bullet before the collision.

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okpalawalter8

Answer:

[tex]v=15.9499m/s[/tex]

Explanation:

From the question we are told that:

Mass of wood [tex]m=1kg[/tex]

force constants [tex]k= 200N-m[/tex]

Coefficient of kinetic friction [tex]\mu= 0.2[/tex]

Bullet mass [tex]m_b= 20 \approx 0.02kg[/tex]

Spring compresion [tex]y=15cm \approx 0.15 m[/tex]

Generally the equation for kinetic energy of bullet [tex]K>E_b[/tex] is mathematically given by

Complete Question

[tex]K.E_b=spring potential energy+work done against friction[/tex]

[tex]K.E_b=\frac{1}{2} mbv^2[/tex]

[tex]\frac{1}{2} m_b v^2=\frac{1}{2} ky^2+\mu my[/tex]

[tex]\frac{1}{2} (0.02)v^2=\frac{1}{2} (0.2)(0.15)^2+0.2(1)(0.15)[/tex]

[tex]v=15.9499m/s[/tex]

[tex]v\approx16m/s[/tex]

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