A 5 kg block sits on a surface with a maximum static coefficient of friction of

Question

A 5 kg block sits on a surface with a maximum static coefficient of friction of 0.60. A 10g bullet is fired into the block. Assume the collision is inelastic. Delta t = 1 second

a) What must be the speed of the bullet to just make the block start moving?

b) If the speed of the bullet is 5 times the speed satisfying part (a), how far will the block slide along the sirface before stopping?

c) How much kinetic energy is dissipated in the bullet/block collision?

d) now assume a completely elastic collision between the block and a 20g "super-ball". What must be speed of the projectile to just move the block?

Explanation / Answer

friction force on the block = 0.6*5*9.8 = 29.5 N

=> 0.01*v/1 = 29.5 => v = 2950m/s

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let block slide x meters

5.01 v ' = 2950*0.01*4 => v ' = 5.9 *4m/s

=> work done by friction = change in kinetic energy

=> 29.5*x = 0.5*5*16*5.9^2

=> x = 47.2 m

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kinetic enrgy dissipated = 0.5 (0.01*16*2950^2 - 5*16*5.9^2) = 694.8 KJ

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