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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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Answer:


v=15.9499m/s

Step-by-step explanation:

From the question we are told that:

Mass of wood
m=1kg

force constants
k= 200N-m

Coefficient of kinetic friction
\mu= 0.2

Bullet mass
m_b= 20 \approx 0.02kg

Spring compresion
y=15cm \approx 0.15 m

Generally the equation for kinetic energy of bullet
K>E_b is mathematically given by

Complete Question


K.E_b=spring potential energy+work done against friction


K.E_b=(1)/(2) mbv^2


(1)/(2) m_b v^2=(1)/(2) ky^2+\mu my


(1)/(2) (0.02)v^2=(1)/(2) (0.2)(0.15)^2+0.2(1)(0.15)


v=15.9499m/s


v\approx16m/s

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