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.

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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