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happycreature
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Hi everyone, this is my first post to the forums. Nice to meet you all and thank you in advance.
A bullet of mass 6.00g is fired horizontally into a wooden block of mass 1.21kg resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.210. The bullet remains embedded in the block, which is observed to slide a distance 0.260m along the surface before stopping.
What is the initial velocity of the bullet?
Mass of bullet = 6.00g = 0.006kg
Mass of block = 1.21kg (+bullet) = 1.216
Coeff. of friction µk = 0.210
∆d = 0.260m
∆KE + ∆PE = -Wnc
KE = 1/2mv^2
Work = F∆d
F(kf) = µk(N)
*F∆t = m(vf - vi)
F(sf) = Coeff. friction x normal force = (.210)(1.21)(9.8) = 2.49 N
Work = F∆d
Work done by friction = (2.49N)(0.260m) = -0.647 J
No potential energy here so...
KEf - KEi = -0.647
KEf = 0 (the block slid to a stop)
KEi = -0.647
1/2mvi^2 = -0.647
vi = √(2)(0.647)/0.006 = 14.64 m/s
...Which is incorrect.
*Because this problem is in the chapter on momentum it seems like they want me to use an equation involving momentum, but I don't know 1)if this problem can or cannot be solved with the work-energy theorem and 2) when I would use the momentum equation in the above scenario.
For the record I also tried finding the initial velocity of the block (1.03 m/s) and putting that as my KEf for the bullet but the difference was negligible.
Any advice greatly appreciated - thank you!
Homework Statement
A bullet of mass 6.00g is fired horizontally into a wooden block of mass 1.21kg resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.210. The bullet remains embedded in the block, which is observed to slide a distance 0.260m along the surface before stopping.
What is the initial velocity of the bullet?
Mass of bullet = 6.00g = 0.006kg
Mass of block = 1.21kg (+bullet) = 1.216
Coeff. of friction µk = 0.210
∆d = 0.260m
Homework Equations
∆KE + ∆PE = -Wnc
KE = 1/2mv^2
Work = F∆d
F(kf) = µk(N)
*F∆t = m(vf - vi)
The Attempt at a Solution
F(sf) = Coeff. friction x normal force = (.210)(1.21)(9.8) = 2.49 N
Work = F∆d
Work done by friction = (2.49N)(0.260m) = -0.647 J
No potential energy here so...
KEf - KEi = -0.647
KEf = 0 (the block slid to a stop)
KEi = -0.647
1/2mvi^2 = -0.647
vi = √(2)(0.647)/0.006 = 14.64 m/s
...Which is incorrect.
*Because this problem is in the chapter on momentum it seems like they want me to use an equation involving momentum, but I don't know 1)if this problem can or cannot be solved with the work-energy theorem and 2) when I would use the momentum equation in the above scenario.
For the record I also tried finding the initial velocity of the block (1.03 m/s) and putting that as my KEf for the bullet but the difference was negligible.
Any advice greatly appreciated - thank you!
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