If the coefficient of friction between block and surface is 0.251 , what was the speed of the bullet immediately before impact?
Answer in units of m/s.
1 answer:
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Refer to the diagram shown below.
The basket is represented by a weightless rigid beam of length 0.78 m.
The x-coordinate is measured from the left end of the basket.
The mass at x=0 is 2*0.55 = 1.1 kg.
The weight acting at x = 0 is W₁ = 1.1*9.8 = 10.78 N
The mass near the right end is 1.8 kg.
Its weight is W₂ = 1.8*9.8 = 17.64 N
The fulcrum is in the middle of the basket, therefore its location is
x = 0.78/2 = 0.39 m.
For equilibrium, the sum of moments about the fulcrum is zero.
Therefore
(10.78 N)*(0.39 m) - (17.64 N)*(x-0.39 m) = 0
4.2042 - 17.64x + 6.8796 = 0
-17.64x = -11.0838
x = 0.6283 m
Answer: 0.63 m from the left end.
The basket is represented by a weightless rigid beam of length 0.78 m.
The x-coordinate is measured from the left end of the basket.
The mass at x=0 is 2*0.55 = 1.1 kg.
The weight acting at x = 0 is W₁ = 1.1*9.8 = 10.78 N
The mass near the right end is 1.8 kg.
Its weight is W₂ = 1.8*9.8 = 17.64 N
The fulcrum is in the middle of the basket, therefore its location is
x = 0.78/2 = 0.39 m.
For equilibrium, the sum of moments about the fulcrum is zero.
Therefore
(10.78 N)*(0.39 m) - (17.64 N)*(x-0.39 m) = 0
4.2042 - 17.64x + 6.8796 = 0
-17.64x = -11.0838
x = 0.6283 m
Answer: 0.63 m from the left end.