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A 50 g bullet moving horizontally at a velocity of 400 m/s collides and embeds itself in a 25 kg block that rests on a rough horizontal surface and is attached to a spring with a spring constant of 180 N/m. The impact causes a compression of 12 cm in the spring. What is the coefficient of friction between the block and the surface?

Respuesta :

onyebuchinnaji

The coefficient of friction between the block and the surface is determines as 0.1.

Final velocity of the block and the bullet

Apply the principle of conservation of linear momentum;

0.05(400) = v(0.05 + 25)

20 = 25.05v

v = 0.8 m/s

Coefficient of friction

Apply the principle of conservation of energy

P.E  - K.E = Ux

μmgx - ¹/₂mv²  = ¹/₂kx²

μ(25.05)(9.8)(0.12) - 0.5(25.05)(0.8)² = 0.5(180)(0.12)

29.459μ - 8.016   = 10.8

29.459μ = 2.784

μ = 0.1

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