The acceleration of gravity is 9.8 m/s 2. After impact, the block slides 9.15 m before coming to rest. 84 g 10.4 g vbullet 94.4 g 9.15 m µ = 0.251 If the coefficient of friction between block and surface is 0.251, what was the speed of the bullet immediately before impact?

Question

Physics

Dillon Cobb

12 October, 05:33

The acceleration of gravity is 9.8 m/s 2. After impact, the block slides 9.15 m before coming to rest. 84 g 10.4 g vbullet 94.4 g 9.15 m µ = 0.251 If the coefficient of friction between block and surface is 0.251, what was the speed of the bullet immediately before impact?

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Answers (1)

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Jase Lindsey · Answer 1

Answer:speed of bullet = 14.1m/s

Explanation:

Vo = (m+m) / m sqrt (2gdu)

d=9.15m

g=9.8m/s^2

u=0.251

Vo = 84+94.4/84 sqrt (2*9.8*9.15*0.251)

Vo=178.4/84 sqrt (45.01)

Vo=2.1*6.71

Vo = 14.1m/s